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

\(=\left(2x^2+6x-x-3\right)-\left(3x^2-4x-6x+8\right)+5x\)

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

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

\(=x^2+20x-11\)

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

\(=10x^3-15x^2+5x-2x\left(x^2-2x+x-2\right)\)

\(=10x^3-15x^2+5x-2x^3+4x^2-2x^2+4x\)

\(=8x^3-13x^2+9x\)

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

\(=3x^2+3x+2x+2-2x^2-6x-2x+1\)

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

\(C=\left(3x+2\right)\left(x-1\right)-2x\left(x+3\right)-2x+1\\ C=3x^2-x-2-2x^2-6x-2x+1\\ C=x^2-9x-1\)

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

\(=x^2-2x+1-x^2-4x-4\)

\(=-6x-3\)

a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6                

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

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                        

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6       => 2x = -10        => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)     

  => (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7             

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = 9  => x = 0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0

=> 3(6x² - 5x +1) - (18x² - 29x + 3) = 0

=> (18x² -15x + 1) -(18x² - 29x +3) = 0

=> 18x² - 15x +1 -18x² + 29x - 3 = 0

=> 14x = 0  => x = 0

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

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

x²+3x+2x+6-x²-5x+2x+10=6

(x²-x²)+(3x+2x-5x+2x)+(10+6)=6

2x+16=6

2x=6-16

2x=-10

x=-10/2

x=-5. Vậy x=-5

b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6

6x²+27x+4x+18-6x²-x-12x-2=7

(6x²-6x²)+(27x+4x-x-12x)+(18-2)=7

18x+16=7

18x=7-16

x=-9/18=-1/2. Vậy x=-1/2

c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0

(9x-3)(2x-1)-(2x-3)(9x-1)=0

9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0

18x²-9x-6x+3-18x²+2x+27x-3=0

(18x²-18x²)+(27x+2x-6x-9x)+(3-3)=0

14x=0

x=0/14

x=0. Vậy x=0

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

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                          

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6  => 2x = -10    => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)

=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = -9  => x = -0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0

=> 3(6x² - 5x + 1) - (18x² - 29x + 3) = 0

=> 18x² - 15x + 3 - 18x² + 29x -3 = 0

=> 14x = 0  => x = 0.

dài quá bạn ơi viết từng câu thôi

a) Ta có: \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[5x-5\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x-5x+10\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

mà \(10\ne0\)

nên x+1=0

hay x=-1

Vậy: x=-1

b) Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)=4\)

\(\Leftrightarrow4x^2-8x+x-2-2x+3-4=0\)

\(\Leftrightarrow4x^2-9x-3=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}-\frac{129}{16}=0\)

\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{129}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{9}{4}=\frac{\sqrt{129}}{4}\\2x-\frac{9}{4}=-\frac{\sqrt{129}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{9+\sqrt{129}}{4}\\2x=\frac{9-\sqrt{129}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{129}}{8}\\x=\frac{9-\sqrt{129}}{8}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{9+\sqrt{129}}{8};\frac{9-\sqrt{129}}{8}\right\}\)

c) Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x^2-9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

mà \(2\ne0\)

nên \(\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;3\right\}\)

d) Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x-2=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

Vậy: x=-1

e) Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+8\right)=3-3x^2\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=12\)

hay x=4

Vậy: x=4

f) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-1\)

\(\Leftrightarrow6x^2-\left(6x^2-4x+15x-10\right)+1=0\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10+1=0\)

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

\(\Leftrightarrow-11x=-11\)

hay x=1

Vậy: x=1

a,\(\Leftrightarrow\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)-17=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x-17=0\)

\(\Leftrightarrow9x-10=0\)

\(\Leftrightarrow x=\frac{10}{9}\)

b,\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

c,\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+1-15=0\)

\(\Leftrightarrow45x=14\)

\(\Leftrightarrow x=\frac{14}{45}\)

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

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

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

\(-5x=3\)

\(x=\frac{-3}{5}\)

tự ghi dấu suy ra ở đằng trước nhé

b) Vì \(2x\left(5-3x\right)=2x\left(3x-5\right)-3\left(x-7\right)=3\)

nên chỉ cần giải: \(6x^2-10x-3x+21=3\)

\(\Leftrightarrow6x^2-13x+21=3\)

\(\Leftrightarrow6x^2-13x+18=0\)

\(\Rightarrow\)pt vô nghiệm

a) <=> \(2x^2-8x+3x-12+x^2-7x+10=3x^2-5x-12x+20\)

<=> \(2x^2-8x+3x-12+x^2-7x+10-3x^2+5x+12x-20=0\)

<=> \(5x-22=0\)

<=> \(5x=22\)

<=> \(x=\frac{22}{5}\)

b) <=> \(24x^2-9x+16x-6-4x^2-7x-16x-28=10x^2+5x-2x-1\)

<=> \(24x^2-9x+16x-6-4x^2-7x-16x-28-10x^2-5x+2x+1=0\)

<=> \(10x^2-19x-33=0\)

<=> \(10x^2-30x+11x-33=0\)

<=> \(10x\left(x-3\right)+11\left(x-3\right)=0\)

<=> \(\left(x-3\right)\left(10x+11\right)=0\)

<=> \(x=3;x=-\frac{11}{10}\)