Cho a+b+c =2015 and \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=\frac{1}{10}\)
Tính \(S=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\)
Cho a+b+c=2015 và\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\) Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}\)
\(\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=\frac{2015}{90}\)
\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2015}{90}\)
\(1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2015}{90}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{2015}{90}-3=\frac{349}{18}\)
a) Cho a + b +c = 2015 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)
Tính S = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
b) cho 2 số a,b thỏa mãn điều kiện a+b=1.Chứng minh a3 +b3 +ab lớn hơn hoặc bằng \(\frac{1}{2}\)
\(a)\) Ta có :
\(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\left(a+b+c\right).\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{a+b+c}{2015}\)
\(\Leftrightarrow\)\(1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=\frac{2015}{2015}\)
\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1-3\)
\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)
Vậy ...
Cho \(\hept{\begin{cases}a+b+c=2001\\\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\end{cases}}\) Tính S = \(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}\)
\(S=\left(\frac{c}{a+b}+1\right)+\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)-3\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
\(=2001\cdot\frac{1}{10}-3=\frac{1971}{10}\)
cho a,b,c là ba số thực khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)chung minh:
\(\frac{1}{a^{2015}}+\frac{1}{b^{2015}}+\frac{1}{c^{2015}}=\frac{1}{a^{2015}+b^{2015}+c^{2015}}\)
cho 3 số a, b, c thuộc R t/m \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a^{^{ }}+b+c}\)
cmr: \(\frac{1}{a^{2015}}+\frac{1}{b^{2015}}+\frac{1}{c^{2015}}=\frac{1}{a^{2015}+b^{2015}+c^{2015}}\)
mong được mọi người giúp đỡ
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{\left(a+b+c\right)c}\)
\(\Leftrightarrow\left(a+b\right)\left(a+b+c\right)c=-\left(a+b\right)ab\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Tự làm nốt
1)Cho a,b,c là các số thực thỏa mãn: a+b+c=2015 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2015}\).Tính \(\frac{1}{a^{2015}}+\frac{1}{b^{2015}}+\frac{1}{c^{2015}}\)
2)Cho n là số dương.Chứng minh:
T= \(2^{3n+1}-2^{3n-1}+1\) là hợp số.
3)Cho a,b,c là ba số dương và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\).Tìm Max A=\(\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ac+a^2}}\)
Cho a+b+c=2014 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2014}\).Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)
Cho a+b+c=2001 và\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)
\(S=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)
\(S=\frac{2001}{b+c}+\frac{2001}{c+a}+\frac{2001}{a+b}-3\)
\(S=2001\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
\(S=2001.\frac{1}{10}-3=\frac{1971}{10}\)
Cho a+b+c=2001 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
1/(a+b) + 1/(b+c) + 1/(c+a)=1/10
<=>(a+b+c)(1/a+b + 1/b+c + 1/c+a)=(a+b+c).1/10
<=>2001.(1/a+b + 1/b+c + 1/c+a)=200,1
<=>2001/a+b + 2001/b+c + 2001/c+a =200,1
<=>a+b+c/a+b + a+b+c/b+c + a+b+c/c+a=200,1
<=>a+b/a+b + c/a+b + b+c/b+c + a/b+c + c+a/c+a + b/c+a
<=>3+ c/a+b + a/b+c + b/c+a=200,1
<=>c/a+b + a/b+c + b/c+a=198,1