cho a,b,c là 3 cạnh 1 tam giác. CMR
A= a/(b+c-a) + b/(a+c-b) +c/(a+b-c) >=3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}=15\)
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}-15=0\)
\(\frac{X-90}{10}-1+\frac{X-76}{12}-2+\frac{X-58}{14}-3+\frac{X-36}{16}-4+\frac{X-15}{17}-5=0\)
\(\frac{X-100}{10}+\frac{X-100}{12}+\frac{X-100}{14}+\frac{X-100}{16}+\frac{X-100}{17}=0\)
\(\left(X-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=> X-100=0
=> X=100
vậy x=100
Ta có: a, b, c là độ dài ba cạnh của tam giác
\(\Rightarrow\hept{\begin{cases}\frac{a}{b+c-a}\\\frac{b}{a+c-b}\\\frac{c}{a+b-c}\end{cases}}>0\)
\(A=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)
\(A+\frac{3}{2}=\frac{a}{b+c-a}+\frac{1}{2}+\frac{b}{a+c-b}+\frac{1}{2}+\frac{c}{b+a-c}+\frac{1}{2}\)
\(A+\frac{3}{2}=\frac{a+b+c}{2\left(b+c-a\right)}+\frac{a+b+c}{2\left(a+c-b\right)}+\frac{a+b+c}{2\left(b+a-c\right)}\)
\(A+\frac{3}{2}=\frac{\left(a+b+c\right)}{2}\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{b+a-c}\right)\)
\(A+\frac{3}{2}\ge\frac{a+b+c}{2}.\frac{9}{b+c-a+a+c-b+b+a-c}\)
\(A+\frac{3}{2}\ge\frac{9}{2}\)
\(\Rightarrow A\ge3\)