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\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\frac{a}{b+c}=\frac{1}{2}\Rightarrow\frac{a}{b}=1\)Bn tự tính phần sau rồi thế vào đẳng thức đó mà tính
KQ: 8
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c=\frac{a}{\frac{2}{1}}=\frac{b}{\frac{3}{2}}=\frac{c}{\frac{4}{3}}\)
Áp dụng TC DTSBN ta có :
\(\frac{a}{\frac{2}{1}}=\frac{b}{\frac{3}{2}}=\frac{c}{\frac{4}{3}}=\frac{a-b}{\frac{2}{1}-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
=> a = 60 ; b = 45 ; c = 40
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
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{\left(b+c+1\right)+\left(a+c+2\right)+\left(a+b-3\right)}{a+b+c}\)
\(=\frac{2.\left(a+b+c\right)}{a+b+c}=2=\frac{1}{a+b+c}\)
\(\Rightarrow a+b+c=\frac{1}{2}\)\(\Rightarrow\hept{\begin{cases}b+c=\frac{1}{2}-a\\a+c=\frac{1}{2}-b\\a+b=\frac{1}{2}-c\end{cases}}\)
Thay vào đề bài ta có: \(\frac{\frac{1}{2}-a+1}{a}=\frac{\frac{1}{2}-b+2}{b}=\frac{\frac{1}{2}-c-3}{c}=2\)
\(\Rightarrow\frac{\frac{3}{2}-a}{a}=\frac{\frac{5}{2}-b}{b}=\frac{\frac{-5}{2}-c}{c}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{3}{2}-a=2a\\\frac{5}{2}-b=2b\\\frac{-5}{2}-c=2c\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3a=\frac{3}{2}\\3b=\frac{5}{2}\\3c=\frac{-5}{2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{5}{6}\\c=\frac{-5}{6}\end{cases}}\)
Vậy \(a=\frac{1}{2};b=\frac{5}{6};c=\frac{-5}{6}\)
ta có \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{b+a}\)
=>\(S+3=3+\left(\dfrac{a}{b+c}+\dfrac{c}{b+a}+\dfrac{b}{c+a}\right)\)
hay \(S+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{b+a}+1\right)\)
=>\(S+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{b+a}\)
=>\(S+3=a+b+c\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
=>\(S+3=2007\cdot\dfrac{1}{90}\)
=>\(S+3=\dfrac{2017}{90}\)
=>S=\(\dfrac{1747}{90}\)
Theo đề: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=\frac{2019}{90}\)
Khai triển:
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
\(=\frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3=\frac{2019}{90}\)
Làm nốt nhé :3