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

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

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

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

Ta có:A =  5x² + y² + z² - 4x - 2xy - z - 1

A = (x² - 2xy + y²) + (4x² - 4x + 1) + (z² - z + 1/4) - 9/4

A = (x - y)² + (2x - 1)² + (z - 1/2)² - 9/4 \(\ge\)- 9/4 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\2x-1=0\\z-\frac{1}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\x=\frac{1}{2}\\z=\frac{1}{2}\end{cases}}\) <=> x =  y = z = 1/2

Vậy MinA = -9/4 khi x = y = z = 1/2

=)) mình cũng làm ntn mà rút gọn ngu -9/4=-3/2 kq sai :v

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 = \(\dfrac{1}{2}\)

câu 1

x^2 -5x +y^2+xy -4y +2014 

=(y^2+xy +1/4x^2) -4(y+1/2x)+4 +3/4x^2-3x+2010

=(y+1/2x-2)^2 +3/4(x^2-4x+4)+2007

=(y+1/2x-2)^2 +3/4(x-2)^2 +2007

GTNN là 2007<=> x=2 và y=1

ta có 

\(x^2+y^2+z^2\)\(=200\)

\(2xy-yz-zx=M\)

\(\Leftrightarrow M+200=x^2+y^2+z^2+2xy-yz-zx\)

\(\Leftrightarrow M+200=\left(x+y\right)^2-z\left(x+y\right)+z^2\)

\(\Leftrightarrow\left(x+y-\frac{z}{2}\right)^2+\frac{3}{4}z^2\ge0\)

\(\Leftrightarrow M\ge-200\)

Điều kiện có 2 nghiệm phân biệt tự làm nha

Theo vi-et ta có:

\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)

\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)

\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)

\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)

Làm nốt nhé

Câu 1:

M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)

=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)

=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)

\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)

\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)

2/ \(S=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=9\)

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) 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

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