Do \(xyz\ne0\) ta có:

\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)

Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)

\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)

Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)

\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)

Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)

\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)

\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=-\dfrac{1}{z^3}\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=0\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}.\left(-\dfrac{1}{z}\right)=0\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)

\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}.xyz\)

\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vậy...

CM bài toán:

nếu a+b+c=0 thì a^3+b^3+c^3=3abc

a^3+b^3+c^3=3abc

=>a^3+b^3+c^3-3abc=0

=>(a+b)^3-3ab(a+b)+c^3-3abc=0

=>[(a+b)^3+c^3]-3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0

vì a+b+c=0 nên a^3+b^3+c^3=3abc

thay a =1/x,b=1/y,c=1/z

áp dụng vào coog thức vừa chứng minh ta đc

\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)

lại có: A=\(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz.\dfrac{3}{xyz}=3\)

vậy................

năm mới vui vẻ

B=\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}) \)

áp dụng đẳng thức câu hỏi https://hoc24.vn/hoi-dap/question/530394.html

đặt \(\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c \)

=> a+b+c=0=>\(a^3+b^3+c^3=3abc\)

=>\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3} =\frac{3}{xyz} \)

=> B=\(\frac{3xyz}{xyz}=3 \)

A sẽ giải thích tại sao đặt được nhân tử vậy cho nhé

Ta có:

\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

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

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

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

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

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Cậu ta làm sai thì làm sao ngonhuminh với thầy phynit hiểu được

1 đó

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Leftrightarrow\left(\dfrac{1}{x}\right)^3+\left(\dfrac{1}{y}\right)^3+\left(\dfrac{1}{z}\right)^3=3.\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}\)

T có: \(M=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\)

\(=xyz.\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\)

\(=xyz.\dfrac{3}{xyz}=3\)