$4 \frac{x}{1 \frac{1}{2 \frac{1}{3}}}=\frac{x}{4 \frac{1}{3 \frac{1}{2}}}$$\Leftrightarrow4 \frac{x}{1 \frac{1}{\frac{7}{3}}}=\frac{x}{4 \frac{1}{\frac{7}{2}}}$\(\Leftrightarrow4 \frac{x}{1 \frac...

#Ngữ văn lớp 1

4

T

Tú

21 tháng 8 2019

1 cách khác nó phức tạp và khó hơn "n" lần :)) Cơ mà nó chẳng khác của cậu là mấy :v

$4+\frac{x}{1+\frac{1}{2+\frac{1}{3}}}=\frac{x}{4+\frac{1}{3+\frac{1}{2}}}$

$\Leftrightarrow4+\frac{x}{1+\frac{1}{\frac{7}{3}}}=\frac{x}{4+\frac{1}{\frac{7}{2}}}$

$\Leftrightarrow4+\frac{x}{1+\frac{3}{7}}=\frac{x}{4+\frac{2}{7}}$

$\Leftrightarrow4+\frac{x}{\frac{10}{7}}=\frac{x}{\frac{30}{7}}$

$\Leftrightarrow4+x.\frac{7}{10}=x.\frac{7}{30}$

$\Leftrightarrow4+\frac{7x}{10}=\frac{7x}{30}$

$\Leftrightarrow120+21x=7x$

$\Leftrightarrow120=7x-21$

$\Leftrightarrow120=-14x$

$\Leftrightarrow-\frac{120}{14}=-\frac{60}{7}=x$

$\Rightarrow x=-\frac{60}{7}$

ZC

zZz Cool Kid zZz

21 tháng 8 2019

Tuấn Huỳnh cách của a có khác gì cách của e đâu.chỉ một bên chọn MSC còn a thì chuyển vế thôi mà

1 tìm x biết ;a, 0-|x + 1| = 5 b, 2 - | $\frac{3}{4}$- x | = $\frac{7}{12}$c, 2 | $\frac{1}{2}$x - $\frac{1}{3}$| - $\frac{3}{2}$= $\frac{1}{4}$d, | x - $\frac{1}{3}$| = $\frac{5}{6}$e, $\frac{3}{4}$- 2 | 2x - $\frac{2}{3}$| =...

I

ImAmNoob!!!

18 tháng 4 2021

1 tìm x biết ;a, 0-|x + 1| = 5 b, 2 - | $\frac{3}{4}$- x | = $\frac{7}{12}$c, 2 | $\frac{1}{2}$x - $\frac{1}{3}$| - $\frac{3}{2}$= $\frac{1}{4}$d, | x - $\frac{1}{3}$| = $\frac{5}{6}$e, $\frac{3}{4}$- 2 | 2x - $\frac{2}{3}$| =...

2

AH

Akai Haruma

Giáo viên

20 tháng 4 2021

d,

$|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.$

e,

$\frac{3}{4}-2|2x-\frac{2}{3}|=2$

$\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}$

$\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0$ (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f,

$\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)$

$\Leftrightarrow 6x-3=10+6x$

$\Leftrightarrow 13=0$ (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

Đúng(1)

AH

Akai Haruma

Giáo viên

20 tháng 4 2021

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

$2-|\frac{3}{4}-x|=\frac{7}{12}$

$|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}$

$\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.$

c,

$2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}$

$2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}$

$|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}$

$\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.$

Đúng(1)

I

ImAmNoob!!!

18 tháng 4 2021

0

MT

Minh Tiến TV

25 tháng 3 2020

giải...

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

25 tháng 3 2020

Bài 1:

ĐKXĐ: x≠1

Ta có: $\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}$

$\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0$

$\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0$

$\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0$

$\Leftrightarrow3x^2-3x=0$

$\Leftrightarrow3x\left(x-1\right)=0$

Vì 3≠0

nên $\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0$

Vậy: x=0

Bài 2:

ĐKXĐ: x≠2; x≠3; $x\ne\frac{1}{2}$

Ta có: $\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}$

$\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0$

$\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0$

$\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0$
$\Leftrightarrow x^2+x-12-x^2-2x+8=0$

$\Leftrightarrow-x-4=0$

$\Leftrightarrow-x=4$

hay x=-4(tm)

Vậy: x=-4

Bài 3:

ĐKXĐ: x≠1; x≠-1

Ta có: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)$

$\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x-\frac{3x\left(x-1\right)}{x+1}$

$\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0$

$\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0$

$\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0$

$\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0$

$\Leftrightarrow-6x^2+10x=0$

$\Leftrightarrow2x\left(-3x+5\right)=0$

Vì 2≠0

nên $\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.$

Vậy: $x\in\left\{0;\frac{5}{3}\right\}$

Bài 4:

ĐKXĐ: x≠1; x≠-3

Ta có: $\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}$

$\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0$

$\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0$

$\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0$

$\Leftrightarrow13x-1=0$

$\Leftrightarrow13x=1$

hay $x=\frac{1}{13}$(tm)

Vậy: $x=\frac{1}{13}$

Bài 5:

ĐKXĐ: x≠1; x≠-2

Ta có: $\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}$

$\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0$

$\Leftrightarrow x+2-7\left(x-1\right)-3=0$

$\Leftrightarrow x+2-7x+7-3=0$

$\Leftrightarrow-6x+6=0$

$\Leftrightarrow-6\left(x-1\right)=0$

Vì -6≠0

nên x-1=0

hay x=1(ktm)

Vậy: x∈∅

Bài 6:

ĐKXĐ: x≠4; x≠2

Ta có: $\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}$

$\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0$

$\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0$

$\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0$

$\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0$

$\Leftrightarrow x^2+x-6+x^2-5x+4+2=0$

$\Leftrightarrow2x^2-4x=0$

$\Leftrightarrow2x\left(x-2\right)=0$

Vì 2≠0

nên $\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0$

Vậy: x=0

Bài 7:

ĐKXĐ: x≠1; x≠-2; x≠-1

Ta có: $\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}$

$\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0$

$\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0$

$\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0$

$\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0$

$\Leftrightarrow-6x^2+13x+8=0$
$\Leftrightarrow-6x^2+16x-3x+8=0$

$\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0$

$\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}$

NT

Nguyễn Thành Trương

25 tháng 3 2020

\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)

Còn lại tương tự mà làm nhé!

TD

Trang Đinh Huyền

22 tháng 8 2019

...

7

MM

minh man

22 tháng 8 2019

1,$\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}$

$\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}$

$\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}$

$\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}$

$\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}$

$\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}$

$\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}$

$\left(x-\frac{9}{4}\right)=\frac{45}{4}$

$x=\frac{45}{4}+\frac{9}{4}$

$x=\frac{27}{2}$

TD

Trang Đinh Huyền

22 tháng 8 2019

Bước cưối 58/21 minh man viết nhầm nên sai

DT

Duong Thi Nhuong

27 tháng 7 2016

Tìm x: $\frac{3\frac{1}{2}x-4}{6}+\frac{2+\frac{1}{4}x}{3}=1\frac{1}{4}x\left(x-1\right)-\frac{7-\frac{3}{4}x}{3}$

#Toán lớp 7

1

H

haphuong01

27 tháng 7 2016

Hỏi đáp Toán

DT

Duong Thi Nhuong

27 tháng 7 2016

ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???

1. So sánh:a. 1 và $\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{50}}$b. $\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{100}}$với $\frac{1}{2}$c. $\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^6}+.....\frac{1}{4^{1000}}$với $\frac{1}{3}$2. Tìm x,...

NA

Ngọc Ánh

7 tháng 8 2016

$A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)$$B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)$

$C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}$

#Toán lớp 5

0

NH

Nguyễn Hoàng Phương Nhi

11 tháng 7 2016

#Toán lớp 7

0

AJ

agelina jolie

6 tháng 6 2016

tìm x biết...

3

PT

Phạm Tuấn Kiệt

6 tháng 6 2016

a) (x + 1/2) . (2/3 − 2x) = 0

$\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.$

$\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.$

$\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.$

b) $\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2$

$\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}$

$\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}$

$\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}$

$\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}$

$\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}$

$\Rightarrow x.\frac{44}{7}=-\frac{8}{7}$

$\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}$

$\Rightarrow x=-\frac{2}{11}$

PT

Phạm Tuấn Kiệt

6 tháng 6 2016

c) $x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}$

$\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}$

$\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}$

$\Rightarrow x.\frac{25}{12}=\frac{25}{12}$

$\Rightarrow x=\frac{25}{12}:\frac{25}{12}$

$\Rightarrow x=1$

d) $5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}$

$\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}$

$\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}$

$\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}$

$\Rightarrow\frac{89}{17}:x=\frac{365}{2002}$

$\Rightarrow x=\frac{89}{17}:\frac{365}{2002}$

$\Rightarrow x\approx28,7$ (số hơi lẻ)

e) $\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}$

$\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}$

$\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}$

$\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.$

$\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.$

$\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.$

Bài 1:Tìm $x\in Z$biết$A=\frac{3}{x-1}$$B=\frac{x-2}{x+3}$$C=\frac{2x+1}{x-3}$Bài 2:Chứng tỏ rằng:$A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}>2$$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1$Bài 3:Tính hợp...

DM

Đỗ Mạnh Huy

24 tháng 7 2020

1

LD

l҉o҉n҉g҉ d҉z҉

24 tháng 7 2020

$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}$

Ta có : $\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}$

$\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}$

...

$\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}$

Cộng vế theo vế

$\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}$

$\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}$

Lại có $\frac{7}{8}< 1$

Theo tính chất bắc cầu => $B< \frac{7}{8}< 1$

$\Rightarrow B< 1\left(đpcm\right)$