\(\left(2x+1\right)^2+\left(4x-2\right)^2-3\left(2x+1\right)\left(4x-2\right)\)

\(=4x^2+4x+1+16x^2-16x+4-6\left(2x+1\right)\left(2x-1\right)\)

\(=20x^2-12x+5-6\left(4x^2-1\right)\)

\(=20x^2-12x+5-24x^2+6=-4x^2-12+11\)

\(\frac{x^2+3x}{x^2+6x+9}+\frac{3}{x-3}+\frac{6x}{9-x^2}\)

ĐKXĐ : x ≠ ±3

\(=\frac{x\left(x+3\right)}{\left(x+3\right)^2}+\frac{3}{x-3}-\frac{6x}{x^2-9}\)

\(=\frac{x}{x+3}+\frac{3}{x-3}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}+\frac{3x+9}{\left(x-3\right)\left(x+3\right)}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\)

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

\(\left(3x+1\right)^2-9\left(x+2\right)^2=-5\)

\(\Leftrightarrow9x^2+6x+1-9\left(x^2+4x+4\right)=-5\)

\(\Leftrightarrow9x^2+6x+1-9x^2-36x-36=-5\)

\(\Leftrightarrow-30x-35=-5\Leftrightarrow-30x=30\Leftrightarrow x=-1\)

a, \(\left(x-y\right)\left(x+y\right)=x^2-y^2\)

\(\Leftrightarrow x^2+xy-xy-y^2=VP\Leftrightarrow x^2-y^2=VP\Leftrightarrow VT=VP\left(đpcm\right)\)

b, \(\left(x+y\right)^2=x^2+2xy+y^2\Leftrightarrow x^2+2xy+y^2=VP\Leftrightarrow VT=VP\left(đpcm\right)\)

c, \(\left(x-y\right)^2=x^2-2xy+y^2\Leftrightarrow x^2-2xy+y^2=VP\Leftrightarrow VT=VP\left(đpcm\right)\)

d, Ta có : \(\left(x+y\right)\left(x^2-xy+y^2\right)=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)

\(=x^3+y^3=VP\Leftrightarrow VT=VP\left(đpcm\right)\)

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

\(=x^4-y^4=VP\Leftrightarrow VT=VP\left(đpcm\right)\)

trả lời nhanh nhanh nhanhhh