\(C=3.\left(x^2-8y^3-15\right)-3\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)

\(=3x^2-24y^3-45-3\left[x\left(x^2+2xy+4y^2\right)-2y\left(x^2+2xy+4y^2\right)\right]\)

\(=3x^2-24y^3-45-3\left[\left(x^3+2x^2y+4xy^2\right)-\left(2x^2y+4xy^2+8y^3\right)\right]\)

\(=3x^2-24y^3-45-3\left(x^3+2x^2y+4xy^2-2x^2y-4xy^2-8y^3\right)\)

\(=3x^2-24y^3-45-3\left(x^3-8y^3\right)\)

\(=3x^2-24y^3-45-3x^3+24y^3\)

\(=3x^2-3x^3-45\)

\(x^3+8y^3+2xy^2+x^2y\)

\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)

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

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

A= \(^{x^3+3x^2y-4xy^2-12y^3=x^2\left(x+3y\right)-4y^2\left(x+3y\right)=\left(x+3y\right)\left(x^2-4y^2\right)}\)

1) 

a) \(2x^2-12x+18+2xy-6y\)

\(=2x^2-6x-6x+18+2xy-6y\)

\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)

\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)

\(=\left(x-3\right)\left(2y+2x-6\right)\)

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

b) \(x^2+4x-4y^2+8y\)

\(=x^2+4x-4y^2+8y+2xy-2xy\)

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

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

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

2)  \(5x^3-3x^2+10x-6=0\)

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

Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)

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

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

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

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

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

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

\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)

\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

Bài làm

a) 2x² - 12x + 18 + 2xy - 6y

= 2x² - 6x - 6x + 18 + 2xy - 6y 

= ( 2xy + 2x² - 6x ) - ( 6y + 6x - 18 )

= 2x( y + x - 3 ) - 6( y + x - 3 )

= ( 2x - 6 ) ( y + x - 3 )

# Học tốt #

Đặt \(A=x^2+2y^2+2xy+2x+4y-1\)

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

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

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

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

Vậy GTNN của \(A\) là \(-3\) khi \(x=0\) và \(y=-1\)

Chúc bạn học tốt ~ 

Đặt \(B=-x^2-2x-y^2-8y-10\)

\(-B=\left(x^2+2x+1\right)+\left(y^2+8y+16\right)-7\)

\(-B=\left(x+1\right)^2+\left(y+4\right)^2-17\ge-17\)

\(B=-\left(x+1\right)^2-\left(y+4\right)^2+17\le17\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}-\left(x+1\right)^2=0\\-\left(y+4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}}\)

Vậy GTLN của \(B\) là \(17\) khi \(x=-1\) và \(y=-4\)

Chúc bạn học tốt ~ 

a ) Sửa lại : \(\left(2x+\dfrac{1}{2}\right)\left(4x^2-x+\dfrac{1}{4}\right)=8x^3+\dfrac{1}{8}\)

b ) Sửa lại : \(\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)=125x^3-8y^3\)

c ) Sửa lại : \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=x^3+8y^3\)

d ) Đ

a)\(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2+2\cdot x\cdot3+9\right)+\left[\left(2y\right)^2-2\cdot2y\cdot3+9\right]=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\2y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{3}{2}\end{matrix}\right.\)

b)\(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-2\cdot x\cdot5+25\right)+\left(y^2-2.y.4+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-5=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=5\\y=4\end{matrix}\right.\)(vô lý)

x⁴-3x³-x+3 = (x⁴-3x³)-(x-3) = x³(x-3)-(x-3) = (x-3)(x³-1) = (x-3)(x-1)(x²+x+1)

3x+3y-x²-2xy-y² = (3x+3y)-(x²+2xy+y²) = 3(x+y)-(x+y)² = (x+y)( 3-x-y)

x²-x-12 = x(x-1)-12

4x⁴+ 4x²y²- 8y⁴

<=> (2x²- 2y²)