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22 tháng 9 2018

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

24 tháng 6 2018
https://i.imgur.com/2COiOAX.jpg
24 tháng 6 2018
https://i.imgur.com/CbuFS19.jpg

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)

`=> (x-3)5 = (2x+1)3`

`=> 5x-15 = 6x+3`

`=> 5x-6x = 15+3`

`=> -x=18`

`=> x=-18`

\(\dfrac{x+1}{22}=\dfrac{6}{x}\)

`=> (x+1)x = 22*6`

`=> (x+1)x = 132`

`=> x^2 + x = 132`

`=> x^2+x-132=0`

`=> (x-11)(x+12)=0`

`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)

\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)

`=> (2x-1)x = 2*5`

`=> 2x^2 - x =10`

`=> 2x^2 - x - 10 =0`

`=> 2x^2 + 4x - 5x - 10 =0`

`=> (2x^2 + 4x) - (5x+10)=0`

`=> 2x(x+2) - 5(x+2)=0`

`=> (2x-5)(x+2)=0`

`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)

`=> (2x-1)(2x+1)=21*3`

`=> 4x^2 + 2x - 2x - 1 = 63`

`=> 4x^2 - 1=63`

`=> 4x^2 - 1 - 63=0`

`=> 4x^2 - 64 = 0`

`=> 4(x^2 - 16)=0`

`=> 4(x^2 + 4x - 4x - 16)=0`

`=> 4[(x^2+4x)-(4x+16)]=0`

`=> 4[x(x+4)-4(x+4)]=0`

`=> 4(x-4)(x+4)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)

`=> (2x+1)(x+1) = 9*5`

`=> (2x+1)(x+1)=45`

`=> 2x^2 + 2x + x + 1 = 45`

`=> 2x^2 + 3x + 1 =45`

`=> 2x^2 + 3x + 1 - 45 =0`

`=> 2x^2+3x-44=0`

`=> 2x^2 + 11x - 8x - 44=0`

`=> (2x^2 +11x) - (8x+44)=0`

`=> x(2x+11) - 4(2x+11)=0`

`=> (x-4)(2x+11)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

15 tháng 6 2023

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)

\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)

 

 

8 tháng 7 2017

Giúp mình nhé các bạn mình đang cần gấp lắm

1 tháng 9 2019

a) 2x-5=3+2x-7x

2x-2x+7x=3+5

7x=8

  x=8/7

vậy x=8/7

1 tháng 9 2019

a) 2x - 5 = 3 + 2x - 7x

=> 2x - 2x + 7x = 3 +5 

=> 7x = 8

=> x = 8/7

b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)

=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)

=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)

\(\Leftrightarrow-2x^2-2x=0\)

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

Vì -2≠0

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

Vậy: x∈{0;-1}

2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)

\(\Leftrightarrow11x-2=0\)

\(\Leftrightarrow11x=2\)

hay \(x=\frac{2}{11}\)

Vậy: \(x=\frac{2}{11}\)

3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

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

\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)

\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)

hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)

Vậy: \(x=\frac{-13}{4}\)

4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)

2 tháng 4 2020

1. \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x=3x^2-x\)

\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)

\(\Leftrightarrow-2x-2x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

2. \(15-5\left(1-2x\right)=12-x\)

\(\Leftrightarrow15-5+10x=12-x\)

\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)

\(\Leftrightarrow-2+11x=0\)

\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)

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

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

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

\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)

\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)

4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)

\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)

\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)

7 tháng 8 2015

2(x - 3) + 5 = 3x - 1

2x-6+5=3x-1

2x-1=3x-1

2x-3x=-1+1

-x=0

x=0

2x(3x + 2) - 5 = 3( 2x^2 - 2x + 1)

6x2+4x-5=6x2-6x+3

6x2+4x-6x2+6x=3+5

10x=8

x=4/5

(3x - 2)(2x - 3) + 5 = 5

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

=>3x-2=0 hoặc 2x-3=0

=>x=2/3 hoặc x=3/2

7 tháng 8 2015

2(x - 3) + 5 = 3x - 1

<=>2x-6+5=3x-1

<=>2x-3x=-1+6-5

<=>-x=0

<=>x=0

 

2x(3x + 2) - 5 = 3( 2x2 - 2x + 1)

<=>6x2+4x-5=6x2-6x+3

<=>4x+6x=3+5

<=>10x=8

<=>x=0,8

 

(3x - 2)(2x - 3) + 5 = 5

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

<=>3x-2=0 hoặc 2x-3=0

<=>x=2/3 hoặc x=3/2

25 tháng 8 2020

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x2 = -144

=> x2 = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x2 - x = 10

=> 2x2 - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x2 - 1 = 63

=> 4x2 = 64

=> x2 = 16

=> x = \(\pm\)4

h) Tương tự

25 tháng 8 2020

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)