\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{-1}{z^3}\)
\(\frac{1}{x^3}+\frac{3}{x^2y}+\frac{3}{xy^2}+\frac{1}{y^3}=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{3}{xyz}\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{3}{xyz}.xyz\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xy}{z^2}+\frac{xz}{y^2}=3\)
khi gấp lên mấy lần thì nó vẫn bằng 0 nên biểu thức đó bằng 0
A= \(\frac{yz}{x^2}\)+\(\frac{xy}{z^2}\)+\(\frac{xz}{y^2}\)= xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\))
Đặt \(\frac{1}{x}\)=a ; \(\frac{1}{y}\)=b; \(\frac{1}{z}\) =c
=> a+b+c=0
Ta có a3+b3+c3- 3abc= (a+b)3+c3-3ab(a+b)-3abc
=(a+b+c)3- 3ab(a+b)-3abc-3(a+b)c(a+b+c)
=(a+b+c)3- 3ab(a+b+c)- 3(a+b)c(a+b+c)
Mà a+b+c=0 => a3+b3+c3- 3abc=0
=>a3+b3+c3=3abc
=> \(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\)=\(\frac{3}{xyz}\)
=> A=xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\))= xyz.\(\frac{3}{xyz}\)=3
(đúng thì tk nha)
1/x + 1/y + 1/z = 0 => 1/x + 1/y = - 1/z
<=> ( 1/x + 1/y )^3 = -1/z^3
1/x^3 + 3/x^2y + 3/xy^2 + 1/y^3 = -1/z^3
<=> 1/x^3 + 1/y^3 + 1/z^3 = -3/x^2y - 3/xy^2 = -3/xy(1/x + 1/y) = 3/xyz
=> xyz = (1/x^3 + 1/y^3 + 1/z^3) =3/xyz . xyz
=> yz/x^2 + xy/z^2 + xz/y^2 = 3
x 1 + y 1 + z 1 = 0⇒ x 1 + y 1 = − z 1 ⇔ x 1 + y 1 3 = z 3 −1 x 3 1 + x 2 y 3 + xy 2 3 + y 3 1 = − z 3 1 ⇔ x 3 1 + y 3 1 + z 3 1 = x 2 y −3 − xy 2 3 = xy −3 x 1 + y 1 = xyz 3 ⇒xyz x 3 1 + y 3 1 + z 3 1 = xyz 3 .xyz ⇒ x 2 yz + z 2 xy + y 2 xz = 3
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}x1+y1+z1=0⇒x1+y1=−z1
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+xz}{xzy}=0\Leftrightarrow xy+yz+xz=0\)
\(A=\frac{yz}{x^2}+\frac{xy}{z^2}+\frac{xz}{y^2}=\frac{x^3y^3+y^3z^3+x^3z^3}{x^2y^2z^2}\)
\(=\frac{x^3y^3+y^3z^3+x^3z^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}\)
\(=\frac{\left(xy+yz+xz\right)\left(x^2y^2+x^2z^2+y^2z^2-xyz\left(x+y+z\right)\right)+3x^2y^2z^2}{x^2y^2z^2}\)
\(=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
vậy \(A=3\)
CÁCH KHÁC:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Leftrightarrow yz+xz+xy=0\left(1\right)\)
\(A=\frac{yz}{^{x^2}}+\frac{xz}{y^2}+\frac{xz}{z^2}=\frac{\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3}{\left(xyz\right)^2}\)
Mặt khác \(\left(1\right)\Leftrightarrow yz+xz=-xy\)
\(\Leftrightarrow\left(yz+xz\right)^3=-xy^3\)
\(\Leftrightarrow\left(yz\right)^3+3xy^2z^3+3x^2yz^3+\left(xz\right)^2+\left(xy\right)^3=0\)
\(\Leftrightarrow\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3=-3xyz^2\left(yz+xz\right)\left(2\right)\)
Vì \(yz+xz=-xy\)nên \(\left(2\right)\Leftrightarrow\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3=3(xyz)^2\)
Thay vào A ta được:
\(A=\frac{3\left(xyz\right)^2}{\left(xyz\right)^2}=3\)
Ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) \(\Rightarrow\) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{z}+\frac{1}{x}=-\frac{1}{y}\end{cases}}\)
lại có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x}+\frac{1}{z}\right)=0\)
hay \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( 1 )
\(A=\frac{yz}{x^2}+\frac{xy}{y^2}+\frac{xz}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\) ( 2 )
Từ (1), (2) \(\Rightarrow A=3\)
đáp án bằng : 3
Cứ động não lên là làm đc đó bạn
