\(M=\left(x^2-2xy+y^2\right)+\left(4x^2-4x+1\right)+\left(z^2-z+\frac{1}{4}\right)-\frac{5}{4}\)

\(M=\left(x-y\right)^2+\left(2x-1\right)+\left(z-\frac{1}{2}\right)^2-\frac{5}{4}>=-\frac{5}{4}\)

=>M min\(=-\frac{5}{4}\)

<=>x=y=z=1/2

Ta có \(C=5x^2+y^2+z^2-4x-2xy-z-1\)

\(=x^2-2xy+y^2+4x^2-4x+1+z^2-z+\dfrac{1}{4}-1-\dfrac{1}{4}-1\)

\(=\left(x-y\right)^2+\left(2x-1\right)^2+\left(z-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)

Ta có \(\left(x-y\right)^2\ge0;\left(2x-1\right)^2\ge0;\left(z-\dfrac{1}{2}\right)^2\ge0\)

=> \(C\ge-\dfrac{9}{4}\)

=> C đạt giá trị nhỏ nhất là \(-\dfrac{9}{4}\) khi

\(\left\{{}\begin{matrix}x-y=0\\2x-1=0\\z-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y=\dfrac{1}{2}\\x=\dfrac{1}{2}\\z=\dfrac{1}{2}\end{matrix}\right.\)

=> \(x=y=z=\dfrac{1}{2}\)

Vậy MinC = \(-\dfrac{9}{4}\)khi x=y=z = \(\dfrac{1}{2}\)

\(M=5x^2+y^2+z^2-4x-2xy-z-1\)

\(=\left(4x^2-4x+1\right)+\left(x^2-2xy+y^2\right)+\left(z^2-z+\dfrac{1}{4}\right)-\dfrac{9}{4}\)

\(=\left(2x-1\right)^2+\left(x-y\right)^2+\left(z-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)

Vậy \(M_{min}=-\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\) ; \(y=\dfrac{1}{2}\)

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Câu 1: 

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

\(\Leftrightarrow x\left(x^2-4\right)-\left(x^3+8\right)=4\)

\(\Leftrightarrow x^3-4x-x^3-8=4\)

\(\Leftrightarrow-4x-8=4\)

\(\Leftrightarrow-4x=12\)

\(\Leftrightarrow x=-3\)

Vậy \(x=-3\)

a,5x(x-1)-3x(x-1)

=(5x-3x)(x-1)

=2x(x-1)

b,4x^2-25

=(2x)^2-5^2

=(2x-5)(2x+5)

c,x^2-x-y^2-y=(x^2-y^2)-(x+y)

= (x-y)(x+y)-(x+y)

=(x+y)(x-y-1)

d, x^2-2xy+y^2-z^2

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

=(x-y-z)(x-y+z)