đặt k là cah hay nhat bn ak

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hay nhể

x-y-z=0 =>x-y=z => 2x - 2y =2z     (1)

x+2y-10z=0 => x+2y =10z             (2)

Cộng 2 vế (1) và (2) : =>3x=12z  => x=4z

Thay x=4z vào x-y-z=0 ta đc:

4z-y-z=0 => 3z-y=0   => y=3z

Thay x=4z;y=3z vào B ta tính đc B=8

hjhj kb vs mik nhé 

, \(B=\frac{2x^2+4xy}{y^2+z^2}=\frac{2x\left(x+2y\right)}{y^2+z^2}\)

\(\hept{\begin{cases}x-y-z=0\\x+2y-10z=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-y=z\\x+2y=10z\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4z\\y=3z\end{cases}}\)

Thay vào B, ta được: \(B=\frac{2.\left(4z\right)^2+4.4z.3z}{\left(3z\right)^2+z^2}=\frac{2.4^2+3.4^2}{3^2+1}=8\)

=> 

 Cho a+b+c=0 và a² +b² +c² =1.Tìm a⁴+b⁴+c⁴.

Lời giải:
$4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0$

$(4x^2+y^2+z^2-4xy-4xz+2yz)+y^2+z^2-6y-10z+34=0$

$(2x-y-z)^2+(y^2-6y+9)+(z^2-10z+25)=0$
$(2x-y-z)^2+(y-3)^2+(z-5)^2=0$

Vì $(2x-y-z)^2\geq 0; (y-3)^2\geq 0; (z-5)^2\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì bản thân mỗi số đó bằng $0$

$\Rightarrow 2x-y-z=y-3=z-5=0$

$\Rightarrow y=3; z=5; x=4$

Khi đó:
$P=0^{2023}+(-1)^{2025}+(5-4)^{2027}=0$

4x² + 2y² + 2z² - 4xy + 2yz - 4xz - 6y - 10z + 34 = 0

<=> [ ( 4x² - 4xy + y² ) - 4xz + 2yz + z² ] + ( y² - 6y + 9 ) + ( z² - 10z + 25 ) = 0

<=> [ ( 2x - y )² - 2( 2x - y )z + z² ] + ( y - 3 )² + ( z - 5 )² = 0

<=> ( 2x - y - z )² + ( y - 3 )² + ( z - 5 )² = 0

\(\hept{\begin{cases}\left(2x-y-z\right)^2\\\left(y-3\right)^2\\\left(z-5\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(2x-y-z\right)+\left(y-3\right)^2+\left(z-5\right)^2\ge0\forall x,y,z\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)

Thế vào S ta được :

S = ( x - 4 )²⁰²⁰ + ( y - 3 )²⁰²⁰ + ( z - 5 )²⁰²⁰

    = ( 4 - 4 )²⁰²⁰ + ( 3 - 3 )²⁰²⁰ + ( 5 - 5 )²⁰²⁰

    = 0 + 0 + 0

    = 0

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

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

=>\(x^2+2xy+y^2=z^2\)

=>\(z^2-xy=x^2+xy+y^2\)

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

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

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

\(=2x^2+5xy+2y^2=(x+2y)(2x+y)\)

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

=>\(x^2+2xz+z^2=y^2\)

=>\(y^2-xz=x^2+xz+z^2\)

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

\(=4yz-\left(y^2+2yz+z^2\right)=-y^2+2yz-z^2=-\left(y-z\right)^2\)

\(yz+2x^2=yz+2\left(y^2+2yz+z^2\right)\)

\(=2y^2+5yz+2z^2=2y^2+4yz+yz+2z^2\)

=2y(y+2z)+z(y+2z)

=(y+2z)(2y+z)

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

=>\(y^2+2yz+z^2=x^2\)

=>\(x^2-yz=y^2+yz+z^2\)

\(4xz-y^2\) =4xz-(x+z)^2

=4xz-\(x^2-2xz-z^2\)

\(=-x^2+2xz-z^2=-\left(x-z\right)^2\)

\(xz+2y^2=xz+2\left(x+z\right)^2\)

\(=xz+2x^2+4xz_{}+2z^2=2x^2+5xz+2z^2\)

\(=2x^2+4xz+xz+2z^2\)

=2x(x+2z)+z(x+2z)

=(x+2z)(2x+z)

2x+y=x+x+y=x-z

\(2y + z = y + (y + z) = y - x = -(x - y)\)

\(2z + x = z + (z + x) = z - y = -(y - z)\)

\(x + 2y = (x + y) + y = -z + y = y - z\)

\(y + 2z = (y + z) + z = -x + z = z - x\)

\(z + 2x = (z + x) + x = -y + x = x - y\)

Ta có: \(A = \frac{4xy - z^2}{xy + 2z^2} \cdot \frac{4yz - x^2}{yz + 2x^2} \cdot \frac{4zx - y^2}{xz + 2y^2}\)

\(=\frac{-(x - y)^2}{(x + 2y)(2x + y)}\cdot\frac{-(y - z)^2}{(y + 2z)(2y + z)}\cdot\frac{-(z - x)^2}{(z + 2x)(2z + x)}\)

\(=\frac{-\left(x-y\right)^2\cdot\left(y-z\right)^2\cdot\left(z-x\right)^2}{(y-z)(x-z)(z-x)[-(x-y)](x-y)[-(y-z)]}\)

\(=\frac{-(x - y)^2 (y - z)^2 (z - x)^2}{-(x - y)^2 (y - z)^2 (z - x)^2}=1\)