Cho a,b,c > 0 thỏa mãn a+b+c = 2019 và \(\frac{1}{a+b}\)+\(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)= 10 . Tính giá trị của P = \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)
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13 tháng 7 2016
a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
- TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
- TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
b) Đề bài sai ^^
S
30 tháng 10 2018
\(a+b=c+\frac{1}{2019}\Leftrightarrow a+b-c=\frac{1}{2019}\Leftrightarrow\frac{1}{a+b-c}=2019\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=2019\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{a+b-c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b-c}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{a+b}{c\left(a+b-c\right)}\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)=\left(a+b\right)ab\)
\(\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ca+bc-c^2-ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a-c\right)-b\left(a-c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-b\right)\left(a-c\right)=0\)
=>a=-b hoặc c=b hoặc a=c
không mất tính tổng quát, giả sử a=-b, ta có:
\(P=\left(-b^{2019}+b^{2019}-c^{2019}\right)\left(-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)=\left(-c\right)^{2019}\cdot\left(\frac{-1}{c}\right)^{2019}=1\)
tương tư với các trường hợp khác ta cũng có P=1
Vậy P=1
VT
15 tháng 9 2017
đặt \(a+b=x,b+c=y;c+a=z\)
ta có \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\Rightarrow3-\frac{1}{x+1}-\frac{1}{y+1}-\frac{1}{z+1}=1\) \(\)
=> \(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1\)
=> \(\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{x}{x+1}=\frac{1}{x+1}\)
Áp dụng bđt cô si ta có \(\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
=> \(\frac{1}{x+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
tương tự ta có
\(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\)
\(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
nhân từng vế của 3 bđt cùng chièu ta có
\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge8\sqrt{\frac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}=8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
=> \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
NH
18 tháng 3 2020
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
30 tháng 8 2019
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Ta có : \(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(\Rightarrow P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{c+a}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(\Rightarrow P+3=2019.10\)
\(\Rightarrow P+3=20190\)
\(\Rightarrow P=20190-3\)
\(\Rightarrow P=20187\)
Vậy P = 20187