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Vậy \(3x^2-5x^2+9x-15=\left(3x-5\right)\left(x^2+3\right)\)

b

\(\left(x+1\right)\left(x-2\right)-x\left(x-3\right)=0\)

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

\(\Leftrightarrow2x-2=0\)

\(\Leftrightarrow x=1\)

b

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

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

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

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=-1;x=-3\)

\(x^4+x^2y^2+y^4\)

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

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

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

\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)

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

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

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

\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)

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

b

Xét hơi bị nhiều TH nhá:(

Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)

TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)

\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)

Bạn tự xét nốt nhá!

c

\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)

\(\Rightarrow x=11;x=3\)

Thay vào .....