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Có
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\\ \Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\\ \Rightarrow a=b=c=d\)
Vậy
\(M=\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\\ =\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\\ =\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}\\ =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\\ =\frac{1+1+1+1}{2}\\ =\frac{4}{2}=2\)
Vậy M=2
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)\(\Rightarrow\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2\left(a+b+c\right)}{a+b+c}\)(1)
Ta có: \(M=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow M=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-abc}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow\)Biểu thức (1) bằng 2
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)\(\Rightarrow M=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy \(M=-1\)hoặc \(M=8\)
a, Ta có : \(\left(abc\right)^2=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)
=> \(abc=\frac{3}{5}\)
Mà ab = 3/5
=> c = 1.
=> \(\left\{{}\begin{matrix}b=\frac{4}{5}\\a=\frac{3}{4}\end{matrix}\right.\)
Vậy ...
b, Ta có : \(a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=36\)
=> \(\left(a+b+c\right)^2=36\)
=> a + b + c = 6.
=> \(\left\{{}\begin{matrix}a=-\frac{12}{6}=-2\\b=\frac{18}{6}=3\\c=\frac{30}{6}=5\end{matrix}\right.\)
Vậy ...
Dễ mà bạn!
Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b+a-b}{a+c+a-c}=\frac{2a}{2a}=1\)
\(\Rightarrow\hept{\begin{cases}a+b=a+c\\a-b=a-c\end{cases}\Leftrightarrow}b=c\)
Ta có: \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{20b^2}{5b^2}=\frac{20}{5}=4\)
Cách khác:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}\Leftrightarrow\left(a+b\right).\left(a-c\right)=\left(a-b\right).\left(a+c\right)\)
\(\Leftrightarrow a^2-ac+ab-bc=a^2+ac-ab-bc\Leftrightarrow-ac+ab=ac-ab\Rightarrow2ac=2ab\Rightarrow b=c\)(vì a.c khác 0)
\(A=\frac{10.c^2+9c^2+c^2}{2c^2+c^2+2c^2}=\frac{20c^2}{5c^2}=4\)