\(\left(3x-2\right)-5\left(2-3x\right)=0\)

\(\Leftrightarrow\left(3x-2\right)+5\left(3x-2\right)=0\)

\(\Leftrightarrow6\left(3x-2\right)=0\Leftrightarrow x=\frac{2}{3}\)

\(2x^2-2x\left(x-3\right)=30\)

\(\Leftrightarrow2x^2-2x^2+6x=30\Leftrightarrow6x=30\Leftrightarrow x=5\)

Ta có:

\(a^2\left(a+1\right)+2a\left(a+1\right)=a\left(a+2\right)\left(a+1\right)\)

Vì a nguyên nên a(a+1)(a+2) là tích của 3 số nguyên liên tiếp

Mà 3 số nguyên liên tiếp chia hết cho 2 và 3 hay chia hết cho 6

Vậy .............

Bài làm 

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

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

Bai lam 

\(x^2y+xy^2+xy=xy\left(x+y+1\right)\)

Hoc tot 

Ta có : \(N=\frac{2}{x-2}.\left(x^3-x^2-2x\right)=\frac{2x}{x-2}\left(x^2-2x+x-2\right)\)

             \(=\frac{2x}{x-2}\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)

a, \(5x-20y=5\left(x-4y\right)\)

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

c, \(x^2-2xy-y^2-16=-\left(x+y\right)^2-16=-\left(x+y-4\right)\left(x+y+4\right)\)

a) ĐKXĐ :  \(x\ne-3;x\ne2\)

b) \(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-9}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)

Để \(A\inℤ\Rightarrow x-3⋮x-2\)

=> \(x-2-1⋮x-2\)

Vì \(x-2⋮x-2\)

=> \(1⋮x-2\)

=> \(x-2\inƯ\left(1\right)\)

=> \(x-2\in\left\{1;-1\right\}\)

=> \(x\in\left\{3;1\right\}\)

Vậy \(x\in\left\{3;1\right\}\)là giá trị cần tìm

a + b , ĐK \(x\ne2;-3\)

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

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5}{\left(x+3\right)\left(x-2\right)}\)

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