\(8x^2y-18y\)

\(=2y\left(x^2-9\right)\)

\(=2y\left(x-3\right)\left(x+3\right)\)

\(3x^3+6x^2+3x-12xy^2\)

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

\(=3x\left[\left(x+1\right)^2-\left(2y\right)^2\right]\)

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

Rút gọn

\(\frac{x^2}{x^2-4}+\frac{1}{x+2}+\frac{1}{x-2}\)(1)

ĐK : \(x\ne\pm2\)

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

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

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

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

\(=\frac{x}{x-2}\)

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

\(\Leftrightarrow2\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow2\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

\(x^2+4y^2-4x-4y+5=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

\(x^2+4y^2-4x-4y+5=0\) 

<=> \(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)

<=> \(\left(x-2\right)^2+\left(2y-1\right)^2=0\)

<=> \(\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

học tốt

Vì \(3< x< 5\)

\(\Rightarrow x=4\)

Ta có : \(C=x^2-2x-5\)

\(=x^2-2x.1+1^2-1^2-5\)

\(=x^2-2x.1+1-1-5\)

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

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

\(\Leftrightarrow\left(x-1\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2-6\ge6\)

Vậy C đạt GTNN <=> x=1

\(4x^2-4x-y^2-8y-15\)

\(=\left(4x^2-4x+1\right)-\left(y^2+8y+16\right)\)

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

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

\(4x^2-4x-y^2-8y-15\)

\(=\left(4x^2-4x-1\right)-\left(y^2-8y-16\right)\)

\(=\left(4x^2-4x-1\right)-\left(y^2+8y+16\right)\)

\(=\left[\left(2x\right)^2-4x-1\right]-\left[y\left(y+8\right)\right]+16\)

\(=\left(2x^2-6x\right)\left(y+12\right)\left(y-4\right)\)

...

Chia hết giấy

Dư là x + 3

Áp dụng hằng đẳng thức hiệu 2 bình phương .

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

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

\(=\left(2x+3y\right)\left(2x-3y\right)\).

_ Chúc bạn học tốt _