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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
\(x^2+y^2-z^2=x^2+\left(y-z\right)\left(y+z\right)=x^2-x\left(y-z\right)=x\left(x-y+z\right)=x\left(-y-y\right)=-2xy\)
Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)
Cộng VTV:
\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)
Lời giải:
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
Suy ra \(yz=-xy-xz\)
\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)
\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)
\(\Rightarrow \frac{yz}{x^2+2yz}=\frac{yz}{(x-z)(x-y)}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)
\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy \(A=0.\)
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: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
<=> \(\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=0\)
<=> 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...