x²=yz  => \(\frac{x}{y}=\frac{z}{x}\)

\(z^2=xy\Rightarrow\frac{z}{x}=\frac{y}{z}\)

\(\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)

áp dụng ... ta có

\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}=\frac{x+z+y}{y+x+z}=1\)

\(\frac{x}{y}=1\Rightarrow x=y\)

\(\frac{z}{x}=1\Rightarrow z=x\)

=>x=y=z

Ta có x2=yz nên x/y=z/x(1)

y2=xz nên x/y=y/z(2)

z2=xy nên z/x=y/z(3)

Từ 1,2,3 suy ra x/y=z/x=y/z(4)

áp dụng t/c dãy tỉ số bằng nhau vào 4 có

x/y=z/x=y/z=x+y+z/x+y+z

vì x, y,z khác 0 nên x+y+z Khác 0

suy ra x+y+z/z+x+y=1

suy ra x/y=z/x=y/z=1

suy ra x=y; x=z; y=z

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Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

 Ta có :x2 = yz , y2 = xz , z2 = xy

=> x2.y2.z2=yz.xz.xy

=>x2.y2.z2=y2.z2.x2

=>xyz=yxz

=> x=y=z

vãi bạn xyz=yxz đã => x=y=z rồi

Bạn kia giải sai rồi!! \(xyz=yxz\) thì chắc gì \(x=y=z?\)

Giải:

Cộng các đẳng thức trên với nhau ta được:

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

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Mà: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Do đó dấu "=" xảy ra khi \(\Leftrightarrow x=y=z\)

Vậy \(x=y=z\) (Đpcm)

Lời giải:

$(x-y)^2\geq 0$ 

$\Leftrightarrow x^2+y^2\geq 2xy$

$\Leftrightarrow 2(x^2+y^2)\geq (x+y)^2$
$\Leftrightarrow 2\geq (x+y)^2$

$\Leftrightarrow \sqrt{2}\geq x+y\geq -\sqrt{2}$

Ta có đpcm.

Bạn mới bổ sung câu b thì làm như sau:

Áp dụng BĐT Cô-si cho các số dương:

$\frac{1}{x}+\frac{1}{y}\geq \frac{2}{\sqrt{xy}}$

$\frac{1}{y}+\frac{1}{z}\geq \frac{2}{\sqrt{yz}}$

$\frac{1}{z}+\frac{1}{x}\geq \frac{2}{\sqrt{zx}}$

Cộng theo vế và thu gọn:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}$

Dấu "=" xảy ra khi $x=y=z$

Ta có: x²=yz,y²=xz,z²=xy

=>x²+y²+z²=yz+xz+xy

=>2x²+2y²+2z²=2xy+2yz+2xz

=>2x²+2y²+2z²-2xy-2yz-2xz=0

=>(2x²-2xy)+(2y²-2yz)+(2z²-2xz)=0

=>(x²-2xy+x²)+(y²-2yz+y²)+(z²-2xz+z²)=0

=>(x-y)²+(y-z)²+(z-x)²=0

Ta thấy : (x-y)²>0 với mọi x,y

(y-z)²>0 với mọi y,z

(z-x)²>0 với mọi x,z

=>(x-y)²+(y-z)²+(z-x)²>0 với mọi x,y,z

Mà (x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

=>x-y=y-z=z-x=0

=>x=y=z

Bạn tham khảo ở đây nhé.

Câu hỏi của Trịnh Hương Quỳnh - Toán lớp 7 - Học toán với OnlineMath

\(x^2=yz,y^2=xz,z^2=xy\Rightarrow x^2+y^2+z^2=xy+yz+zx\Leftrightarrow2x^2+2y^2+2z^2=2xy+2xz+2y\Leftrightarrow\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y,y=z,x=z\Leftrightarrow x=y=z\)

\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-xz=b\\z^2-xy=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\) \(\Rightarrow ax+by+cz=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)⋮\left(x+y+z\right)\)

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