ta có :

\(A=-\left(x^2-x+\frac{1}{4}\right)-\left(2y^2-y+\frac{1}{8}\right)-\left(4z^2-z+\frac{1}{16}\right)+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(\frac{7}{16}-\left(x-\frac{1}{2}\right)^2-2\left(y-\frac{1}{4}\right)^2-4\left(z-\frac{1}{8}\right)^2\le\frac{7}{16}\)

Vậy GTLN của A là \(\frac{7}{16}\text{ khi }\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{4}\\z=\frac{1}{8}\end{cases}}\)

x^2y^2−2x^2y+2xy^2

ta có :

\(\hept{\begin{cases}\left(x+y\right)^2=x^2+2xy+y^2\text{ và }&2\left(x+y\right)\left(x-y\right)=2x^2-2y^2&\end{cases}}\)

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

\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)

\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)

\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)

Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).

Vậy \(minA=1\)khi \(x=y=1,z=3\).

\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)

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

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)

\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)

Dấu '' = '' xảy ra khi:

\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}z=0\\y=1\\x=1\end{cases}}\)

a, \(\left(x-5\right)^2-16=0\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\Leftrightarrow x=1;x=9\)

b, \(\left(2x-5\right)^2=49=7^2\)

TH1 : \(2x-5=7\Leftrightarrow x=6\)

TH2 : \(2x-5=-7\Leftrightarrow x=-1\)

c, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

d, \(9x^2-6x=-1\Leftrightarrow9x^2-6x+1=0\Leftrightarrow\left(3x-1\right)^2=0\Leftrightarrow x=\frac{1}{3}\)

phân tích thành nhân tử: x(x-y)³-y(y-x)²-y²(x-y)

x(x-y)³-y(y-x)²-y²(x-y)

= (y-x)(y^2-2y-3x)

nah bạn chúc bạn học tốt nha 

x(x-y)³ -y(y-x)² - y²(x-y)

= x(x-y)³ - y(x-y)² - y²(x-y)

= (x-y)[x(x-y)² - y(x-y) - y²]

=(x-y) [x(x² - 2xy + y²) - xy + y² -y²]

=(x-y)(x³ -2x²y +y²x - xy)

=x(x-y)(x² - 2xy + y² -y) 

=x(x-y)[(x-y)²-y]

c) 4x(x-3) - (2x+1)² =12

4x² - 12x - (4x² + 4x +1) = 12

4x² -12x -4x² - 4x -1 =12

(4x² - 4x²) - (12x + 4x) -1 =12

-16x = 13

x= -13/16

Vậy x = -13/16

c, \(4x\left(x-3\right)-\left(2x+1\right)^2=12\)

\(\Leftrightarrow4x^2-12x-4x^2-4x-1=12\)

\(\Leftrightarrow-16x-1=12\Leftrightarrow x=-\frac{13}{16}\)