1/x(x-y)(x-z)+1/y(y-z)(y-x)+1/z(z-x)(z-y)
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15 tháng 9 2019
a ) \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
= \(\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
= \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{-1}{\left(x-z\right)\left(y-z\right)}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)
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17 tháng 7 2023
\(\left(x+y+z\right).\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{2017}{672}\)
\(\Rightarrow\left(\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}\right)=\dfrac{2017}{672}\)
\(\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}=\dfrac{2017}{672}\)
\(\Rightarrow3+\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}\)
\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}-3=\dfrac{2017}{672}-\dfrac{2016}{672}=\dfrac{1}{672}\)
\(\Rightarrow C=\dfrac{1}{672}\)
GH
19 tháng 6 2023
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
NN
17 tháng 8 2020
d) \(\left(a^2+a\right)^2+4\left(a^2+a\right)-12=\left(a^2+a\right)^2+4\left(a^2+a\right)+16-4\)
\(=\left(a^2+a+2\right)^2-4=\left(a^2+a+2-4\right)\left(a^2+a+2+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+6\right)=\left(a-1\right)\left(a+2\right)\left(a^2+a+6\right)\)