\(a,\)

\(x^3y-y\)

\(=y\left(x^3-1\right)\)

\(=y\left[\left(x-1\right)\left(x^2+x+1\right)\right]\)

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

\(b,\)

\(x^3y+y\)

\(=y\left(x^3+1\right)\)

\(=y\left[\left(x+1\right)\left(x^2-x+1\right)\right]\)

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

\(c,\)

\(\left(x-y\right)^2-x\left(y-x\right)\)

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

\(=\left(x-y\right)\left(x-y+x\right)\)

\(=\left(x-y\right)\left(2x-y\right)\)

\(D=x^2+5y^2-2xy+4y+3\)

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

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

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

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)

Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)

\(a,\left(8x^3-1\right):\left(2x-1\right)\)

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

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

+ Thương: \(4x^2+2x+1\)

+ Dư: \(0\)

\(\sqrt{33-20\sqrt{10}}+\dfrac{4}{\sqrt{2}}\)

\(\approx\sqrt{-30,25}+2\sqrt{2}\)

Vì \(\sqrt{...}\ge0\)

Mà \(\sqrt{-30,25}< 0\) (vô lí)

\(\Rightarrow\sqrt{33-20\sqrt{10}}+\dfrac{4}{\sqrt{2}}=\varnothing\)

Ta có: \(BC^2=AB^2+AC^2\) (định lí pitago)

\(\Rightarrow122^2=\left(\dfrac{5AC}{6}\right)^2+AC^2\) (do \(\dfrac{AB}{AC}=\dfrac{5}{6}\))

\(\Rightarrow AC=12\sqrt{61}cm\) (do \(AC>0\))

Ta có: \(AC^2=CH.BC\) (hệ thức lượng trong tam giác vuông)

\(\Rightarrow CH=\dfrac{AC^2}{BC}=\dfrac{\left(12\sqrt{61}\right)^2}{122}=72cm\)

\(\Rightarrow y=72cm\)

Lại có: \(BH+HC=BC\)

\(\Rightarrow BH=BC-HC=122-72=50cm\)

\(\Rightarrow x=50cm\)

\(a,\)

\(\left[2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right]:\left(x-y\right)\)

\(=\left[2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right]:\left(x-y\right)\)

\(=\left\{\left(x-y\right)\left[2\left(x-y\right)^2-7\left(x-y\right)+1\right]\right\}:\left(x-y\right)\)

\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)

\(b,\)

\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:\left[5\left(x-y\right)^2\right]\)

\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)

\(\sqrt{x^2+1}+2=0\)

\(\Leftrightarrow\sqrt{x^2+1}=-2\)

Vì \(\sqrt{x^2+1}\ge0\forall x\in R\)

Mà \(\sqrt{x^2+1}=-2\) (vô lí)

Vậy phương trình vô nghiệm.

\(a,\)

\(A+B+C\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-\left(5x^3y+3x^2y^2+17y^4+1\right)\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)

\(=\left(16x^4-15x^4\right)+\left(-9y^4-6y^4-17y^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)-1\)

\(=x^4-32y^4-10x^3y-x^2y^2-1\)

\(b,\)

\(A-C+B=A+B-C=x^4-32y^4-10x^3y-x^2y^2-1\)

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Rightarrow24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1-33\)

\(\Rightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1-33\)

\(\Rightarrow24x^2-4x^2-10x^2+16x-9x-16x-7x+2x-5x=6+28-1-33\)

\(\Rightarrow10x^2-19x=0\)

\(\Rightarrow x\left(10x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\dfrac{19}{10}\right\}\)

\(\sqrt{9-3\sqrt{8}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}+\sqrt{5-2\sqrt{6}}-\sqrt{2-\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

\(=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{6}+\sqrt{2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{6}-\sqrt{3}\right|-\sqrt{6}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|\)

\(=\sqrt{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}+\sqrt{3}-\sqrt{2}\) (do \(\sqrt{6}-\sqrt{3}>0;\sqrt{3}-\sqrt{2}>0\))

\(=0\)