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\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}\)
\(\Leftrightarrow\frac{b}{a}+\frac{c}{a}=\frac{a}{b}+\frac{c}{b}=\frac{a}{c}+\frac{b}{c}\)
Do đó \(P=\left(\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}\right)=3\left(\frac{b}{a}+\frac{c}{a}\right)=\frac{3\left(b+c\right)}{a}\)
Ta làm như sau:
\(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)
=\(\frac{6}{3.6}\)+\(\frac{6}{6.9}\)+\(\frac{6}{9.12}\)+...\(\frac{6}{30.33}\)
=2 (\(\frac{3}{3.6}\)+\(\frac{3}{6.9}\)+\(\frac{3}{9.12}\)+...+\(\frac{3}{30.33}\)
=2 (\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))
=2 ( \(\frac{1}{3}-\frac{1}{33}\))
=2.\(\frac{10}{33}\)=\(\frac{2.10}{33}\)=\(\frac{20}{33}\)
\(\frac{6}{18}+\frac{6}{54}+\frac{6}{108}+...+\frac{6}{990}\)
=\(\frac{6}{3.6}+\frac{6}{6.9}+\frac{6}{9.12}+...+\frac{6}{30.33}\)
= 2.(\(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\))
=2.(\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))
=2.[\(\frac{1}{3}+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{30}+\frac{1}{30}\right)+\frac{-1}{33}\)]
=2.\(\left[\frac{1}{3}+\frac{-1}{33}\right]\)
=2.\(\left[\frac{11}{33}+\frac{-1}{33}\right]\)
=2.\(\frac{10}{33}\)
=\(\frac{20}{33}\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
Từ dãy tỉ số bằng nhau đó, ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
hay \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4\)
Do đó, \(\frac{a+b+c+d}{a}=4\) => a=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{b}=4\) =>b=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{c}=4\) =>c=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{d}=4\) => d=\(\frac{a+b+c+d}{4}\)
=>a=b=c=d
a+bc+d
Do đó, M=\(\frac{a+b}{c+d}+\frac{b+c}{c+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
Vậy M có giá trị là 4
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(=>Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=>Q=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{a+c}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(=>Q=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(=>Q=259.15-3=3882\)
Vậy Q=3882
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{259-\left(b+c\right)}{b+c}+\frac{259-\left(a+c\right)}{a+c}+\frac{259-\left(a+b\right)}{a+b}\)
\(=259.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)+\left[\frac{-\left(b+c\right)}{b+c}+\frac{-\left(a+c\right)}{a+c}+\frac{-\left(a+b\right)}{a+b}\right]\)
tới đây tự làm tiếp
Ta có: 4/9<a/b
=>4b<9a hay 5a+4a>2b+2b
5a-2b>4a+2b
3>4a+2b(1)
Ta có: a/b<10/21
=>21a<10b hay 5a+16a<2b+8b
5a-2b<8b-16a(2)
Từ (1);(2) =>4a+2b<8b-16a
4a+16a<8b-2b
20a<6b
a/b<6/20
Vậy a/b<6/20 thì thỏa mãn đề*nghĩ v*
bình phương 2 vế của 1/a + 1/b +1/c =2 ta đk:
1/a^2 +1/b^2 + 1/c^2 + 2 x (a+b+c) / abc =4
1/a^2 + 1/b^2 + 1/c^2 +2 =4
=> 1/a^2 + 1/b^2 + 1/c^2 =2
olm đã có 
so sánh hả
ko biết nữa