\(\left(2+x\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x+2\right)^2\left(x-2\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left[\left(x+2\right)^2-x^2-2x-4\right]=\left(x-2\right)\left(x^2+4x+4-x^2-2x-4\right)=\left(x-2\right)2x=2x^2-4x\)

\(\left(x+2\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

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

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

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

\(=x^3-8-x^3-5\)

=-13

\(\left(x-2\right)\cdot\left(x^2+2x+4\right)-\left(x^3+5\right)\\ =x^2-8-x^3-5\\ =-13\)

phần dưới là tìm x

\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)

a) Ta có: \(\left(2x+3\right)^2-4\left(x-2\right)\left(x+2\right)\)

\(=4x^2+12x+9-4\left(x^2-4\right)\)

\(=4x^2+12x+9-4x^2+16\)

\(=12x+25\)

b) Ta có: \(\dfrac{x+6}{x^2-4}-\dfrac{2}{x\left(x+2\right)}\)

\(=\dfrac{x\left(x+6\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+4x+4}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x+2}{x\left(x-2\right)}\)

\(\dfrac{x+1}{x^2-4}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x+1}=x-2\)

làm giúp mình câu 22 với 
 

\(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)\)

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

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

\(=x^6-64\)

\((x+4)(x-2)-(x-3)^2\)

\(=(x⋅x+x⋅(−2)+4⋅x+4⋅(−2))−(x⋅x+x⋅(−3)+(−3)⋅x+(−3)⋅(−3))\)

\(=(x^2-2x+4x-8)-(x^2-3x-3x+9)\)

\(\) \(=(x^2+2x-8)-(x^2-6x+9)\)

\(=x^2+2x-8-x^2+6x-9\)

\(=8x−17\)

(x+4)(x-2)-(x-3)^2

=x^2-2x+4x-8-x^2+6x-9

=2x-8-9

=2x-17

tichs ddi bn owi

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

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

\(=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]\left[\left(x+2\right)\left(x^2-2x+4\right)\right]\)

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

\(=x^6-64\)

\(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)\)

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

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

\(=x^6+64\)

\(\begin{array}{l}(x - 4) + \left[ {({x^2} + 2x) + (7 - x)} \right]\\ = x - 4 + ({x^2} + 2x + 7 - x)\\ = x - 4 + {x^2} + 2x + 7 - x\\ = {x^2} + (x + 2x - x) + ( - 4 + 7)\\ = {x^2} + 2x + 3\end{array}\)