Tính:a) $\dfrac{{3{a^2}}}{{10{b^3}}} \cdot \dfrac{{15b}}{{9{a^4}}}$ b) \(\dfrac{{x - 3}}{{{x^2}}} \cdot \dfrac{{4x}}{{{x^2}

1

H9

HT.Phong (9A5)

CTVHS

23 tháng 7 2023

a) $\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}$

$=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}$

$=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}$

b) $\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}$

$=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}$

$=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}$

$=\dfrac{4}{x\left(x+3\right)}$

c) $\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}$

$=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}$

$=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}$

$=\dfrac{2\left(a-3\right)}{a}$

d) $\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)$

$=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}$

$=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}$

$=x+\dfrac{2-x^2}{x\left(x-1\right)}$

Đúng(1)

VL

Vui lòng để tên hiển thị

23 tháng 7 2023

=))) để r xem

B

Buddy

23 tháng 7 2023

Tính:

a) $\dfrac{{4{x^2} + 2}}{{x - 2}} \cdot \dfrac{{3x + 2}}{{x - 4}} \cdot \dfrac{{4 - 2x}}{{2{x^2} + 1}}$

b) $\dfrac{{x + 3}}{x} \cdot \dfrac{{x + 2}}{{{x^2} + 6x + 9}}:\dfrac{{{x^2} - 4}}{{{x^2} + 3x}}$

Thực hiện các phép tính sau:a) $\dfrac{{8y}}{{3{x^2}}} \cdot \dfrac{{9{x^2}}}{{4{y^2}}}$b) $\dfrac{{3x + {x^2}}}{{{x^2} + x + 1}} \cdot \dfrac{{3{x^3} - 3}}{{x + 3}}$c) $\dfrac{{2{x^2} + 4}}{{x - 3}} \cdot \dfrac{{3x + 1}}{{x - 1}}:\dfrac{{{x^2} + 2}}{{6 - 2x}}$d) $\dfrac{{2{x^2}}}{{3{y^3}}}:\left( { - \dfrac{{4{x^3}}}{{21{y^2}}}} \right)$e) $\dfrac{{2x + 10}}{{{x^3} - 64}}:\dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x - 8}}$f) \(\dfrac{1}{{x + y}}\left( {\dfrac{{x + y}}{{xy}} - x -...

1

AD

@DanHee

23 tháng 7 2023

$a,=\dfrac{2\left(2x^2+1\right).\left(3x+2\right).2\left(2-x\right)}{\left(x-2\right)\left(x-4\right)\left(2x^2+1\right)}=\dfrac{-4.\left(3x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-4\left(3x+2\right)}{x-4}\\ b,=\dfrac{\left(x+3\right).\left(x+2\right)}{x.\left(x+3\right)^2}\times\dfrac{x\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+3\right)\left(x+2\right)x\left(x+3\right)}{x.\left(x+3\right)^2.\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}$

Đúng(1)

B

Buddy

23 tháng 7 2023

1

TN

Thảo Nguyễn

23 tháng 7 2023

$a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x $

$c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}$

$d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}$

$e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}$

$f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1$

Đúng(1)

NN

Nguyễn Ngọc Thanh Trúc

10 tháng 11 2017

a $\dfrac{25x^3y}{7z}\cdot\dfrac{28z}{15x^2y^5}$

b $\dfrac{x^2+3x+9}{2x+10}\cdot\dfrac{x+5}{x^3+27}$

c $\dfrac{3x-6}{x-1}\cdot\dfrac{1-x^3}{10-5x}$

d $\dfrac{3x-2}{x^2+1}\cdot\dfrac{x-1-x^2}{4-9x^2}$

4

NN

Nguyễn Nam

11 tháng 11 2017

Nguyễn Ngọc Thanh Trúc đề là gì

NN

Nguyễn Ngọc Thanh Trúc

11 tháng 11 2017

thực hiện phép tính

Tính giá trị các biểu thức sau theo cách hợp lí nhất. a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$; b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}}...

TD

Thầy Đức Anh

Giáo viên VIP

27 tháng 4 2022

#Toán lớp 7

12

PT

Phạm Trường Giang

VIP

22 tháng 6 2022

a) $A = [\frac{2}{7} (\frac{1}{4} - \frac{1}{3})] : [\frac{2}{7} (\frac{1}{3} - \frac{2}{5})] = (\frac{1}{4} - \frac{1}{3}) : (\frac{1}{3} - \frac{2}{5}) = 1 \frac{1}{4}$ .
b) $B = \frac{\frac{3}{4} (\frac{1}{5} - \frac{2}{7} - \frac{1}{3} + \frac{2}{7})}{\frac{1}{5} (\frac{2}{7} + \frac{1}{3}) - \frac{1}{3} (\frac{2}{7} + \frac{1}{3})} = \frac{\frac{3}{4} (\frac{1}{5} - \frac{1}{3})}{(\frac{1}{5} - \frac{1}{3}) (\frac{2}{7} + \frac{1}{3})} = 1 \frac{11}{52}$ .

T

TINH

13 tháng 7 2022

a) $\mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}$ .
b) $\mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}$

BV

Biển Vũ Đức

26 tháng 1 2019

a,

$\dfrac{x^4+x^2+1}{x^2}=\dfrac{x^2+x+1}{x}$

b,$3\cdot\left(\dfrac{x+3}{x-2}\right)^2+68\cdot\left(\dfrac{x-3}{x+2}\right)^2-46\cdot\dfrac{x^2-9}{x^2-4}=6$

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

14 tháng 1 2023

a: $\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}$

$\Leftrightarrow\dfrac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2}=\dfrac{x^2+x+1}{x}$

$\Leftrightarrow\dfrac{x^2-x+1}{x^2}=\dfrac{1}{x}$

=>x^2=x(x^2-x+1)

=>x(x-x^2+x-1)=0

=>x(-x^2+2x-1)=0

=>x=0(loại) hoặc x=1(nhận)

b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2

=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)

=>$x\simeq28,4$

TL

Tuyết Ly

14 tháng 5 2022

Thực hiện phép...

1

NL

Nguyễn Lê Phước Thịnh

14 tháng 5 2022

a: $=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}$

b: $=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}$

$=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}$

$=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}$

c: $=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}$

=1/3

d: $=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)$

$=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}$

Đúng(4)

HK

Hoàng Khánh Linh

16 tháng 3 2022

e)$\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}$

f)$\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}$
g)
$\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}$
Mn giúp em với ạ

#Toán lớp 6

1

HT

Hồ Tuyết Anh

VIP

16 tháng 3 2022

$e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}$

$=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}$

$=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}$

$f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}$

$=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}$

$=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}$

$g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}$

$=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)$

$=\dfrac{5}{9}\cdot1=\dfrac{5}{9}$

Đúng(3)

O

★彡✿ทợท彡★

5 tháng 5 2022

a) $\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}$

b) $\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)$

câu này làm cx đc hoặc ko làm cx ko sao :)

$\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}$

#Toán lớp 6

2

2611

5 tháng 5 2022

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

Đúng(5)

NN

Nguyễn Ngọc Khánh Huyền

5 tháng 5 2022

a) ^{_{$\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}$}}

^{_{$=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}$}}

^{_{$=-\dfrac{5}{24}$}}

b) ^{_{$\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)$}}

^{_{$=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)$}}

_{^{$=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}$}}

_{^{$=\dfrac{6}{7}\cdot\dfrac{1}{3}$}}

_{^{$=\dfrac{2}{7}$}}

_{^{$\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}$}}

_{^{$=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}$}}

_{^{$=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}$}}

_^$=1$

Đúng(4)

NM

Nguyễn Mary

22 tháng 12 2017

CMR

a, $A=\dfrac{2^3+1}{2^3-1}\cdot\dfrac{3^3+1}{3^3-1}\cdot\dfrac{4^3+1}{4^3-1}\cdot....\cdot\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}$

b, $B=\dfrac{2^3-1}{2^3+1}\cdot\dfrac{3^3-1}{3^3+1}\cdot...\cdot\dfrac{n^3-1}{n^3+1}>\dfrac{2}{3}$

1

ND

nam do

22 tháng 12 2017

a)Nhận xét

$\dfrac{n^3+1}{n^3-1}=\dfrac{\left(n+1\right)\left(n^2-n+1\right)}{\left(n-1\right)\left(n^2+n+1\right)}=\dfrac{\left(n+1\right)\left[\left(n-0,5\right)^2+0;75\right]}{\left(n-1\right)\left[\left(n+0,5\right)^2+0,75\right]}$

Áp dụng công thức trên:

$A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}....\dfrac{9^3+1}{9^3-1}$

$=\dfrac{\left(2+1\right)\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right)\left[\left(2+0,5\right)^2+0,75\right]}.\dfrac{\left(3+1\right)\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right)\left[\left(3+0,5\right)^2+0,75\right]}...\dfrac{\left(9+1\right)\left[\left(9-0,5\right)^2+0,75\right]}{\left(9-1\right)\left[\left(9+0,5\right)^2+0,75\right]}$

$=\dfrac{3\left(1,5^2+0,75\right)}{\left(2,5^2+0,75\right)}.\dfrac{4\left(2,5^2+0,75\right)}{2\left(3,5^2+0,75\right)}...\dfrac{10\left(8,5^2+0,75\right)}{8\left(9,5^2+0,75\right)}$

$=\dfrac{3.4....10}{1.2.....8}.\dfrac{1,5^2+0,75}{9,5^2+0,75}$

$=\dfrac{9.10}{2}.\dfrac{3}{91}$

$=\dfrac{3}{2}.\dfrac{90}{91}< \dfrac{3}{2}$

$\Rightarrowđpcm$

b) Làm tương tự