(x+2)(x+3)-(x-2)(x+5)=0

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

a) x^2+5x+6-x^2+7x-10-6=0

12x-10=0

12x=10

x=5/6

Cậu ơi giúp mình 2 câu dưới nữa ược không?

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

<=>14x=0

<=>x=0

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

<=>2(9x+8)=7

<=>18x+16=7

<=>18x=-9

<=>x=-1/2

TÌM X

a) (3x+2)(2x+9)-(6x+1)(x+2)=7

=> 6x² + 31x +18 - 6x² - 13x - 2 - 7 = 0

=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2

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

=> x² + 3x - 10 - x² - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0

=> x + 5 = 0 => x = -5

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

=> 18x² - 15x +3 - 18x² + 29x -3 = 0 => 14x = 0 => x = 0

a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)

b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)

c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

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

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

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

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

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.

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

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

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1