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)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

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

a: =>(x-3)(x+1)=0

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

c: =>(x-5)(x+1)=0

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

h: =>x^2-4x+4-3=0

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

a) ( x + 2 )( 3 - 4x ) = x² + 4x + 4

<=> ( x + 2 )( 3 - 4x ) = ( x + 2 )²

<=> 3 - 4x = x + 2

<=> -4x - x = 2 - 3

<=> -5x = -1

<=> x = \(\frac{1}{5}\)

b) x(2x - 7) - 4x + 14 = 0

<=> x(2x - 7) = 4x - 14

<=> x(2x - 7) = 2(2x - 7)

<=> x = 2

c) 3x - 15 = 2x(x - 5)

<=> 3(x - 5) = 2x(x - 5)

<=> 3 = 2x

<=> x = \(\frac{3}{2}\)

d) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)

<=> 3x - 2 = 5x - 8

<=> 3x - 5x = -8 + 2

<=> -2x = -6

<=> x = 3

a) ( 5x - 4)(4x + 6)=0

<=> \([^{5x-4=0}_{4x+6=0}< =>[^{x=\frac{4}{5}}_{x=\frac{-6}{4}}\)

Vậy S = \(\left\{\frac{4}{5};\frac{-6}{4}\right\}\)

b) ( 3,5x - 7 )( 2,1x - 6,3 ) = 0

<=> \([^{3,5x-7=0}_{2,1x-6,3=0}< =>[^{x=2}_{x=3}\)

Vậy S = \(\left\{2;3\right\}\)

c) ( 4x - 10 )( 24 + 5x ) = 0

<=> \([^{4x-10=0}_{24+5x=0}< =>[^{x=\frac{5}{2}}_{x=\frac{-24}{5}}\)

Vậy S = \(\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

d) ( x - 3 )( 2x + 1 ) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy S = \(\left\{3;\frac{-1}{2}\right\}\)

e) ( 5x - 10 )( 8 - 2x ) = 0

<=> \(\left[{}\begin{matrix}5x-10=0\\8-2x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy S = \(\left\{2;4\right\}\)

f) ( 9 - 3x )( 15 + 3x ) = 0

<=> \(\left[{}\begin{matrix}9-3x=0\\15+3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{3;-5\right\}\)

Học tốt nhaaa !

3x(x - 10) = x - 10

(x - 10)(3x - 1) = 0

Th1:

x - 10 = 0

x = 10

TH2:

3x - 1 = 0

3x = 1

x = 1/3

Vậy x = 10 hoặc x = 1/3

x(x + 7) - (4x + 28) = 0

x(x + 7) - 4(x + 7) = 0

(x + 7)(x - 4) = 0

Th1:

x + 7 = 0

x = - 7

Th2:

x - 4 = 0

x = 4

Vậy x = - 7 hoặc x = 4

x(x - 4) = 2x - 8

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

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

Th1:

x - 2 = 0

x = 2

Th2:

x - 4 = 0

x = 4 

Vậy x = 2 hoặc x = 4

(2x + 3)(x - 1) + (2x - 3)(x - 1) = 0

(x - 1)(2x + 3 + 2x - 3) = 0

4x(x - 1) = 0

Th1:

x = 0

Th2:

x - 1 = 0

x = 1

Vậy x = 0 hoặc x = 1

a)

\(3x\left(x-10\right)=x-10\)

\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(x-10\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=10\end{array}\right.\)

b)

\(x\left(x+7\right)-\left(4x+28\right)=0\)

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-7\end{array}\right.\)

c)

\(x\left(x-4\right)=2x-8\)

\(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)

d)

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

\(\Rightarrow2\left(2x+3\right)\left(x-1\right)=0\)

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

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