a) bn xem lại xem đề bài có đúng k nhé !

Nếu đúng thì kq sẽ là 1

b) 

\(\Rightarrow x\in\begin{cases}0\\\frac{10}{3}\end{cases}\)

c)

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

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

\(\Leftrightarrow x^3-5x^2-x-3=-4x^2-3x+1\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+4\right)=0\)

=>x-1=0

hay x=1

a,(3+x)(x²-9)-(x-3)(x²+3x+9)

=(3x²-27+x³-9x)-(x³-27)

=3x²-27+x³-9x-x³+27

=3x²-9x

=3x(x-3)

b,(x+6)²-2x(x+6)+(x-6)(x+6)

=x²+12x+36-2x²-12x+x²-36

=0

a) \(\left(3+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(3x^2+x^3-27-9x\right)-\left(x^3-27\right)\)

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

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

b) \(\left(x+6\right)^2-2x\left(x+6\right)+\left(x-6\right)\left(x+6\right)\)

\(=\left(x^2+12x+36\right)-\left(2x^2+12x\right)+\left(x^2-36\right)\)

\(=x^2+12x+36-2x^2-12x+x^2-36\)

\(=0\)

a) 4 + 9( x+6) = -2 + 25 = 23

=> 9(x+6) = 23 -4 =19

=> 9x + 54 = 19

=> 9x = 19 -54 =-35

=> x = \(-\dfrac{35}{9}\)

b) \(\dfrac{5}{x}-\dfrac{1}{8}=\dfrac{1}{2}+\left(-\dfrac{5}{12}\right)\)

\(\dfrac{5}{x}-\dfrac{1}{8}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{1}{12}\)

\(\dfrac{5}{x}=\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{5}{24}\)

\(\Rightarrow x=24\)

c) Đề đúng chứ ?

bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

\(=> 2x+3=7\)

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)