A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)

B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

Câu C) bạn xem lại đề nha mik tính ko đc

D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)

a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7

<=> 3x² - x - 2 - ( 3x² + 7x + 2 ) = 7

<=> 3x² - x - 2 - 3x² - 7x - 2 = 7

<=> -8x - 4 = 7

<=> -8x = 11

<=> x = -11/8

b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8

<=> 12x² + 28x + 15 - ( 12x² + x - 6 ) = 8

<=> 12x² + 28x + 15 - 12x² - x + 6 = 8

<=> 27x + 21 = 8

<=> 27x = -13

<=> x = -13/27 

c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9

<=> 2x² + 6x - ( 2x² + 3x + 1 ) - 5 = 9

<=> 2x² + 6x - 2x² - 3x - 1 - 5 = 9

<=> 3x - 6 = 9

<=> 3x = 15

<=> x = 5

d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10

<=> 20x² - 23x - 21 - ( 20x² - 12x - 27 ) = 10

<=> 20x² - 23x - 21 - 20x² + 12x + 27 = 10

<=> -11x + 6 = 10

<=> -11x = 4

<=> x = -4/11

a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)

b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)

c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)

d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)

a) \(\left(x+2\right)^2-9\left(x-2\right)^2=0\)

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

\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x-4\right)=0\\\left(8-2x\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

b)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)

c) \(\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(2x^2-5x+18\right)\left(8x^2+8x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)

\(a,\left(2x-1\right)^2=49\)

\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(4x^2+28x+49=9x^2+36x+36\)

\(4x^2+28x+49-9x^2-36x-36=0\)

\(-5x^2-8x+13=0\)

\(5x^2+13-5x-13=0\)

\(x\left(5x+13\right)-1\left(5x+13\right)=0\)

\(\left(x-1\right)\left(5x+13\right)=0\)

\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)

\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(x=-5\)

\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)

\(25x^2-30x+9-16x^2+56x-49=0\)

\(9x^2+26x-40=0\)

\(9x^2+36x-10x-40=0\)

\(9x\left(x+4\right)-10\left(x+4\right)=0\)

\(\left(9x-10\right)\left(x+4\right)=0\)

\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)

e)

\(\left(x+2\right)^2=9\cdot\left(x^2-4x+4\right)\\ \Leftrightarrow x^2+4x+4-9x^2+32x-32=0\\ \Leftrightarrow-8x^2+36x-28=0\\ \Leftrightarrow-2x^2+9x-7=0\\ \Leftrightarrow-7+7x+2x-2x^2=0\\ \Leftrightarrow-7\cdot\left(1-x\right)+2x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(-7+2x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-7+2x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{7}{2}\\x=1\end{matrix}\right.\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

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

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

\(x\left(2x-3\right)=0\)

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

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

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

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

* Trả lời:

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

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

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

\(\Leftrightarrow x=12\)

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

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

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

\(\Leftrightarrow33x=28\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow-12x=4\)

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

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a)4(x+2)-7(2x-1)+9(3x-4)=30                                                     b)2(5x-8)-3(4x-5)=4(3x-4)+11

<=>4x+8-14x+7+27x-36=30                                            <=>10x-16-12x+15=12x-16+11

<=>17x-21=30                                                                    <=>  -14x=-4     <=>x=2/7

<=>17x=51

<=>x=3 

phần cvaf d đâu bạn ơi

phần c  hình như có vấn đề

d)(5x-3).4x-2x(10x-3)=15

<=>20x²-12x-20x²+6x=15

<=>   -6x=15

<=>x=-15/6