cho a+b+c=4034 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2}\)
tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
ai giải đc mình tick cho
Cho a+b+c = 4034 ; \(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}=\frac{1}{2}\)
Tính \(\frac{a}{c+b}+\frac{b}{a+c}+\frac{c}{a+b}\)
a+b+c=4034=p
\(\frac{a}{c+b}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{p-\left(c+b\right)}{c+b}+\frac{p-\left(a+c\right)}{a+c}+\frac{p-\left(a+b\right)}{a+b}\)
\(=\frac{p}{c+b}+\frac{p}{a+c}+\frac{p}{a+b}-3=p\left[\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right]-3\)
\(=4034.\frac{1}{2}-3=2017-3=2014\)
a)cho a+b+c=2015. và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)tính A=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
b) cho \(\frac{a}{b}=\frac{c}{d}\). CMR \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+3d^2}{2c^2+3cd}\)
giúp mình với mình tick cho
a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
Thay vào rồi c/m nhé
a) Từ đẳng thức : \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow A+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(\Rightarrow A+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{a+c}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow A+3=2015.\frac{1}{5}\)
\(\Rightarrow A+3=403\)
\(\Rightarrow A=400\)
Vậy A = 400
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó : \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2\left(bk\right)^2-3b^2k+5b^2}{2\left(bk\right)^2+3b^2k}=\frac{2k^2b^2-3b^2k+5b^2}{2b^2k^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(1\right)\);
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2\left(dk\right)^2-3d^2k+5d^2}{2\left(dk\right)^2+3d^2k}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2k^2+3d^2k}=\frac{d^2.\left(2k^2-3k+5\right)}{d^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(2\right)\)
Từ (1) và (2) => \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+5d^2}{2c^2+3cd}\)(đpcm)
Cho a + b + c = 2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=\frac{1}{2017}\)
Tính A = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Mk rất gấp, các bn giúp mk vs!!! Mk sẽ tick cho ai trả lời nhanh và đúng nhất
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
To Kudo :
Cho a,b,c là 3 số dương thỏa mãn : \(a+b+c=\frac{1}{2}\) . CMR:
\(\frac{\frac{1}{2}c+ab}{a+b}+\frac{\frac{1}{2}a+bc}{b+c}+\frac{\frac{1}{2}b+ac}{a+c}\ge1\)
P/s: ko làm đc bảo a để a post lời giải lên cho :) Nhưg a nghĩ e sẽ làm đc !
Giải hộ !
Đặt \(A=\frac{\frac{1}{2}c+ab}{a+b}+\frac{\frac{1}{2}a+bc}{b+c}+\frac{\frac{1}{2}b+ac}{a+c}\)
\(=\frac{\left(a+b+c\right)c+ab}{a+b}+\frac{\left(a+b+c\right)a+bc}{b+c}+\frac{\left(a+b+c\right)b+ac}{a+c}\)
\(=\frac{ac+bc+c^2+ab}{a+b}+\frac{a^2+ab+ac+bc}{b+c}+\frac{ab+b^2+bc+ac}{a+c}\)
\(=\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\)
Áp dụng bđt Cô-si cho 2 số dương :
\(\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\sqrt{\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)}}\)
\(=2\sqrt{\left(a+c\right)^2}\)
\(=2\left(a+c\right)\)
C/m tương tự :
\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
\(\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(b+c\right)\)
Cộng từng vế của 3 bđt trên lại ta được :
\(2A\ge2\left(a+b+b+c+c+a\right)\)
\(\Leftrightarrow2A\ge4\left(a+b+c\right)\)
\(\Leftrightarrow A\ge2\left(a+b+c\right)=2.\frac{1}{2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b=c\\a+b+c=\frac{1}{2}\end{cases}\Leftrightarrow a=b=c=\frac{1}{6}}\)
Vậy .............
1) Cho a, b, c ≠ 0 và a ≠b thỏa mãn a + b + c = 2 và (a2 - bc)(b - abc) = (b2 - ac)(a - abc). Tính S = \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
2) Cho a, b, c > 0. CMR: \(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{a^2+b^2+c^2}{2}\)
Làm được đến đâu thì làm nhé. Ai nhanh và đúng thì mình sẽ tick và add friends nhé. Thanks. Please help me!!!
\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)
\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)
\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)
Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)
\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)
Rồi tương tự các kiểu:v
Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)
\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))
Không phải dùng tới Cauchy-Schwarz:D
mình chưa hiểu?
có thể giải thích rõ hơn đc ko
Cho a+b+c=2014 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2014}\).Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)
CMR nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)và a+b+c=abc thì ta có
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=7\)
các bn giải giúp mình nhé , mình tick cho
Vì \(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=3 ==> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)=9= \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
ta có \(\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)= \(\frac{2\left(a+b+c\right)}{abc}\)=2
==> đpcm
cho a+b+c=2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=> (a+b+c).(1/a+b + 1/b+c +1/c+a) = 2017/90
=> a+b+c/a+b + a+b+c/b+c + a+b+c/c+a = 2017/90
=> 1 + c/a+b + 1 + a/b+c + 1 + b/c+a = 2017/90
=> a/b+c + b/c+a +c/a+b = 2017/90 - 3 = 1747/90
Vậy S = 1747/90
Tk mk nha
Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b
=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1
= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1
= 2010.(1/b+c + 1/c+a + 1/a+b) - 3
= 2010.1/3 - 3 = 667
Vậy S = 667
Tk mk nha
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
Ta có \(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
=\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{2010}{3}=670\)
\(\Rightarrow S=667\)