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b/
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
* \(\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b+c=3a\\2c+a=3b\\2a+b=3c\end{matrix}\right.\)
+)\(\Rightarrow\left\{{}\begin{matrix}c=3a-2b\\a=3b-2c\\b=3c-2a\end{matrix}\right.\)
\(\Rightarrow\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)=abc\left(1\right)\)
+) \(\Rightarrow\left\{{}\begin{matrix}2b=3c-a\\2c=3b-a\\2a=3c-b\end{matrix}\right.\)
\(\Rightarrow\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)=8abc\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{abc}{8abc}=\dfrac{1}{8}\)
\(\Rightarrow P=\dfrac{1}{8}\)
3a) A=\(\dfrac{5}{x+xy+xyz}+\dfrac{5}{y+yz+1}+\dfrac{5xyz}{z+xz+xyz}\)
=\(\dfrac{5}{x\left(1+y+yz\right)}+\dfrac{5}{y+yz+1}+\dfrac{5xy}{1+x+xy}\)
=\(\dfrac{5}{x\left(1+y+zy\right)}+\dfrac{5x}{x\left(1+zy+y\right)}+\dfrac{5xy}{x\left(1+y+zy\right)}\)
=\(\dfrac{5+5x+5xy}{x\left(1+yz+y\right)}\)
=\(\dfrac{5x\left(yz+1+y\right)}{x\left(1+yz+y\right)}=5\)
a/ \(x-y-z=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-z=y\\y-x=-z\\y+z=x\end{matrix}\right.\)
\(\Leftrightarrow\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\left(\dfrac{x}{x}-\dfrac{z}{x}\right)\left(\dfrac{y}{y}-\dfrac{x}{y}\right)\left(\dfrac{z}{z}+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{z+y}{z}\)
\(=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
b/ \(M=\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}\)
\(=\dfrac{3a-b}{2a+\left(a-b\right)}+\dfrac{3b-a}{2b-\left(a-b\right)}\) (do \(a-b=7\))
\(=\dfrac{3a-b}{2a+a-b}+\dfrac{3b-a}{2b-a+b}\)
\(=\dfrac{3a-b}{3a-b}+\dfrac{3b-a}{3b-a}\)
\(=1+1=2\)