Tìm x biết:$(5^x 5^{x 1} 5^{x 2}):31=(3^{2x} 3^{2x 1} 3^{2x 2}):13$ CMR: $\frac{1}{3} \frac{1}{3^2} \frac{1}{3^3} \frac{1}{3^4} ... \frac{1}{3^{2018}} \frac{1}{3^{2019}}-\frac{1}{2}$ là một số âm...

#Toán lớp 7

6

ZC

zZz Cool Kid_new zZz

18 tháng 1 2019

$\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}$

$=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}$

$=2+\frac{2017}{\left|2018x-2019\right|+1}$có giá trị lớn nhất

$\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}$có giá trị lớn nhất

$\Rightarrow\left|2018x-2019\right|+1$có giá trị nhỏ nhất

Mà $\left|2018x-2019\right|\ge0$

$\Rightarrow\left|2018x-2019\right|+1\ge1$

Dấu "=" xảy ra khi và chỉ khi:

$\left|2018x-2019\right|=0$

$\Leftrightarrow x=\frac{2019}{2018}$

Vậy $M_{MAX}=2019$tại $x=\frac{2019}{2018}$

ZC

zZz Cool Kid_new zZz

18 tháng 1 2019

$\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}$

$\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}$

$\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}$

$\Rightarrow5^x=3^{2x}$

Mà $\left(5;3\right)=1$

$\Rightarrow x=2x=0$

RA

♡๛ρɦạ๓ ɦ๏àйǥ ℓąйツ♡

24 tháng 9 2019

tìm x biết: a)$\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}$

b)

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

c)$3-\frac{2}{2x-3}=\frac{2}{5}+\frac{2}{9-6x}-\frac{3}{2}$

#Toán lớp 7

4

C

ctk_new

24 tháng 9 2019

a) Đặt $x-1=a$

$pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}$

$\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}$

$\Leftrightarrow\frac{31}{2}=2$(vô lí)

Vậy pt vô nghiệm

K

KAl(SO4)2·12H2O

24 tháng 9 2019

a) $\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}$

$\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}$

$\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}$

$\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}$

$\frac{31}{2}=2$

=> không có x thỏa mãn đề bài.

b) $\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}$

$\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}$

$\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}$

$\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}$

$10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)$

$7-4x-3x^2=25x-25$

$7-4x-3x^2-25x+25=0$

$32-29x-3x^2=0$

$3x^2+29x-30=0$

$3x^2+32x-3x-32=0$

$x\left(3x+32\right)-\left(3x+32\right)=0$

$\left(3x+32\right)\left(x-1\right)=0$

$\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}$

$\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}$

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 - $\frac{2}{3}$| =...

#Toán lớp 6

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

Bài 1. Giải các phương trình sau 1) $\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}-2x$ 2) $\frac{x-3}{5}=6-\frac{1-2x}{3}$ 3) $2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)$ 4) $\frac{2x+3}{3}=\frac{5-4x}{2}$ 5) $\frac{5x+3}{12}=\frac{1+2x}{9}$ 6) $x-\frac{x+1}{3}=\frac{2x+1}{5}$ 7) $\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6$ 8) $\frac{2\left(3x+1\right)+1}{4}-5=\frac{2 \left(3x-1\right)}{5}-\frac{3x+2}{10}$ 9)...

#Toán lớp 6

0

NT

Nguyễn Thị Thu Hiền

30 tháng 3 2020

0

LD

Lê Đức Hoàng

16 tháng 3 2023

giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 3 2023

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>$x=\dfrac{-3\pm\sqrt{61}}{2}$

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

MT

Minh Tiến TV

25 tháng 3 2020

giải...

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!

Giải phương trình: 1. $\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}$ 2. $\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}$ 3. $\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}$ 4. $\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1$ 5. $\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}$ 6. $\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}$ 7....

NC

Nguyễn Châu Mỹ Linh

8 tháng 1 2020

1

BA

B.Thị Anh Thơ

8 tháng 1 2020

1.

$\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}$

$MC:12$

Quy đồng :

$\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}$

$\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}$

$\Leftrightarrow6x+9-\left(10x+6\right)=3x-4$

$\Leftrightarrow6x+9-3x=-4-9+16$

$\Leftrightarrow-7x=3$

$\Leftrightarrow x=\frac{-3}{7}$

2.$\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}$

$MC:20$

Quy đồng :

$\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}$

$\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)$

$\Leftrightarrow30x+15-20=15x-2$

$\Leftrightarrow15x=3$

$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$

NN

Nguyễn Ngọc Linh

27 tháng 2 2019

a,$\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}$

b,$\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}$

c,$\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}$

2

S

satoshi

27 tháng 2 2019

sorry em mới lớp 7

NN

Nguyễn Ngọc Linh

15 tháng 4 2019

lớp 7 kệ e chị có bắt e tl đâu

DT

ĐẶNG THỊ VIỆT HÀ

3 tháng 4 2019

Bài 2: Tìm x biết

a, $\frac{3}{5}+x=\frac{5}{6}$

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

c, $2\frac{2}{3}.x-8\frac{2}{3}=3\frac{1}{3}$

d, $\frac{5}{13}+2x=\frac{3}{13}$

#Toán lớp 6

1

DV

Đặng Viết Thái

3 tháng 4 2019

a, x=$\frac{7}{30}$