\(\Rightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2acxz\)

\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2bcyz+2acxz\)

\(\Rightarrow\left(a^2y^2-2abxy+b^2x^2\right)+\left(a^2z^2-2acxz+c^2x^2\right)+\left(b^2z^2-2bcyz+c^2y^2\right)=0\)

\(\Rightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)

vì \(\frac{x}{a}=\frac{y}{b}\Rightarrow ay=bc\Rightarrow\left(ay-bx\right)^2=0\)

   \(\frac{y}{b}=\frac{z}{c}\Rightarrow cy=bz\Rightarrow\left(bz-cy\right)^2=0\)

   \(\frac{x}{a}=\frac{z}{c}\Rightarrow cx=az\Rightarrow\left(az-cx\right)^2=0\)

\(\Rightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)luôn đúng

\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

Sửa đề \(x^4+y^4+z^4=\frac{1}{2}\left(x^2+y^2+z^2\right)^2\)

Ta có: \(x+y+z=0\)

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

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

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

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

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

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

\(\Leftrightarrow x^4+y^4+z^4=2x^2y^2+2y^2z^2+2z^2x^2\)

\(\Leftrightarrow2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2=\left(x^2+y^2+z^2\right)^2\)

\(\Leftrightarrow x^4+y^4+z^4=\frac{1}{2}\left(x^2+y^2+z^2\right)^2\left(đpcm\right)\)

32/3 nha ban

thích ra vẻ bằng cách viết tiếng anh à

Đề của toán tiếng anh mà

There are 18 students have neither a dog or a cat

_ Study well _

We have : \(A=\)\(9x^2+8-12\)

=\(\left[\left(3x\right)^2-2.3x.2+4\right]+4\)

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

But \(B=2\)⁽²⁾

From  ⁽¹⁾ and  ⁽²⁾

\(\Rightarrow A>B\)