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Theo t/c dãy tỉ số=nhau:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=>a=b=c=d
Thay vào biểu thức A ,ta đc:
\(A=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Vậy A=2
Vì a/b=1=>a=b;b/c=1=>b=c;c/d=1=> c=d;d/a=1=>a=d
=>a=b=c=d
OK?~_~
Đặt \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)
Không mất tính tổng quát giả sử \(a\ge b\ge c\ge d\)=>\(a^2\ge b^2\ge c^2\ge d^2\)
=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)
=>\(A\le\frac{4}{d^2}\)=>\(d^2\le4\)=>\(d\in\text{ }\text{{}\pm1,\pm2\text{ }\)
Xét \(d=\pm1\)=> vô lí
Xét d=\(\pm\)2=> a=b=c=d=\(\pm\)2
=> M=ab+cd=4+4=8
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Leftrightarrow6a-2a=2b+3b\)
\(\Leftrightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow4a-3a=3b-3c+2b\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a=4a-3c\)
\(\Leftrightarrow3a=3c\)
\(\Rightarrow a=c\)
\(\Rightarrow P=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^5}{\left(8a\right)^2\left(4a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=\frac{8^3}{4^3}=2^3=8\)
Theo đầu bài ta có:
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Do \(a+b+c=259\Rightarrow\hept{\begin{cases}a=259-\left(b+c\right)\\b=259-\left(a+c\right)\\c=259-\left(a+b\right)\end{cases}}\)
Từ đó suy ra:
\(\Leftrightarrow Q=\frac{259-\left(b+c\right)}{b+c}+\frac{259-\left(a+c\right)}{a+c}+\frac{259-\left(a+b\right)}{a+b}\)
\(\Leftrightarrow Q=\left(\frac{259}{b+c}-\frac{b+c}{b+c}\right)+\left(\frac{259}{a+c}-\frac{a+c}{a+c}\right)+\left(\frac{259}{a+b}-\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=\left(259\cdot\frac{1}{b+c}+259\cdot\frac{1}{a+c}+259\cdot\frac{1}{a+b}\right)-\left(\frac{b+c}{b+c}+\frac{a+c}{a+c}+\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=259\cdot\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-\left(1+1+1\right)\)
Do \(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}=15\) nên:
\(\Leftrightarrow Q=259\cdot15-3\)
\(\Leftrightarrow Q=3882\)
a=259-(b+c)
b=259-(c+a)
c=259-(a+b)
Thay vào Q
Q=259-(a+b)/a+b+259-(b+c)/b+c+259-(c+a)/c+a
Q=259/a+b+259/b+c+259/c+a-3
Q=259.(1/a+b+1/c+a+1/b)+c-3
Q=259x15-3
Q=3882
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=1+1+1+1=4\)
Từ tỷ lệ thức:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1.\)do a,b,c,d dương
vậy,
A = 4*1/2 = 2.