ta có 

a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)

\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)

\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)

b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2

\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)

\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)

a)

<=> 10x - 35 + 16x - 10 = 5 

<=> 10x + 16x = 5 + 35 + 10

<=> 26x = 50

<=> x = 50/26 = 25/13

1, x= 2

2, x = 4

**** bạn mình trước nhé

trieu dang sai ket qua vi chua doi dau

1) 5x + 3 = 2(x + 7)

5x + 3 = 2x + 2.7

5x + 3 = 2x + 14

5x - 2x = 14 - 3

3x        = 11

x          = 11/3

Vậy x = 11/3

2) 5x - 25 = 2x - 3

5x - 2x  =  -3 + 25

3x         =  22

x           = 22/3

Vậy x = 22/3

a) \(2.\left|5x-3\right|-2x=14\)

\(2\left|5x-3\right|=14+2x\)

\(\left|5x-3\right|=\frac{14+2x}{2}\)

\(\Rightarrow\orbr{\begin{cases}5x-3=\frac{-14-2x}{2}\\5x-3=\frac{14+2x}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\left(5x-3\right).2=-14-2x\\\left(5x-3\right).2=14+2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}10x-6+2x=-14\\10x-6-2x=14\end{cases}\Rightarrow\orbr{\begin{cases}12x=-14+6\\8x=14+6\end{cases}}}\Rightarrow\orbr{\begin{cases}12x=-8\\8x=20\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)

vậy \(\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)

Những câu sau tương tự nhé. 

làm giúp đi

a) \(M(x) = A(x) + B(x) \\= 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4} \\=(4x^4-4x^4)+(-7x^3+7x^3)+(6x^2-5x^2)+(-5x+5x)+(-6+4)\\= {x^2} - 2.\)

b) \(A(x) = B(x) + C(x) \Rightarrow C(x) = A(x) - B(x)\)

\(\begin{array}{l}C(x) = A(x) - B(x)\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 - ( - 5{x^2} + 7{x^3} + 5x + 4 - 4{x^4})\\ = 4{x^4} + 6{x^2} - 7{x^3} - 5x - 6 + 5{x^2} - 7{x^3} - 5x - 4 + 4{x^4}\\ =(4x^4+4x^4)+(-7x^3-7x^3)+(6x^2+5x^2)+(-5x-5x)+(-6-4)\\= 8{x^4} - 14{x^3} + 11{x^2} - 10x - 10\end{array}\) 

a) \(\left(x-4\right)^2-\left(x-4\right)=0\)

\(\left(x-4\right)\left(x-4-1\right)=0\)

\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)

\(\left(x-7\right)\left(5x^2+7\right)=0\)

\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)

\(x=7\)

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

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

\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)

a) (x - 4)^2=(x - 4)

(x - 4) (x -4)=(x -4 )

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

(x-4) (x-4-1)=0

(x-4) (x-5)=0

TH1:x-4=0                          TH2:x-5=0

            x=4                                      x=5