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\(\left\{{}\begin{matrix}xy+yz+xz=0\\x,y,z\ne0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=0\)\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{y^3}+\dfrac{1}{x^3}=\dfrac{3}{zyz}\)
\(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3xyz}{xyz}=3\)
Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo
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 \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
<=> \(\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=0\)
<=> yz + xz + xy = 0
=> (yz)3 + (xz)3 + (xy)3 = 3x2y2z2
\(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
= \(\dfrac{y^3z^3}{x^2y^2z^2}+\dfrac{x^3z^3}{x^2y^2z^2}+\dfrac{x^3y^3}{x^2y^2z^2}\)
= \(\dfrac{3x^2y^2z^2}{x^2y^2z^2}\)
= 3
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...
a)
\(\dfrac{x^2+x-6}{x^3-4x^2-18x+9}=\dfrac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(=\dfrac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x^2-7x+3\right)\left(x+3\right)}=\dfrac{x-2}{x^2-7x+3}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
ta có : \(xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)
\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=0\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3.\dfrac{1}{x^2}.\dfrac{1}{y}+3.\dfrac{1}{x}.\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{xyz}\)
Do đó : \(xyz.\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)
\(\Leftrightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy giá trị của biểu thức \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)