Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,b,c khác nhau đôi một nghĩa là từng cặp số khác nhau ,là:
+a khác b
+b khác c
+c khác a
\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0\)
Suy ra: \(ab==-\left(bc+ac\right)=-bc-ac\)
\(bc=-\left(ab+ac\right)=-ab-ac\)
\(ac=-\left(ab+bc\right)=-ab-bc\)
Nên \(a^2+2ab=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự,ta cũng có: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\)
\(c^2+2ab=\left(c-a\right)\left(c-b\right)\)
Vậy \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
"Chấm" nhẹ hóng cao nhân ạ :)
P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)
\(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)
Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
Thay vào A ta được :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\frac{c\left(a+b\right)\left(a+b+c\right)}{abc\left(a+b+c\right)}+\frac{ab\left(a+b\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(ab+ac+bc+c^2\right)}{abc\left(a+b+c\right)}=0\)
\(\Rightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Để \(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)thì:
a+b=0 hoặc b+c=0 hoặc a+c=0
\(\Rightarrow\)2 trong 3 số đó phải đối nhau
(a+b+c)2=a2+b2+c2
=>2(ab+bc+ac)=0
=>ab+bc+ac=0
=> bc=-ab-ac
=>\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ac-ab+bc}\)=\(\frac{a^2}{\left(a-c\right)\left(a-b\right)}\)
Tuong tu => \(\frac{b^2}{b^2+2ac}=....\)
\(\frac{c^2}{c^2+2ab}=...\)
=> \(\frac{a^2}{a^2+2bc}+....\)=\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)+...
=\(\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
=1
CM theo chiều ngược lại , nếu a ; b ; c là 3 cạnh tam giác
thì tổng các phân thức trên > 1 ( 1 )
\(\frac{a^2+b^2-c^2}{2ab}+1=\frac{\left(a+b\right)^2-c^2}{2ab}\) ; \(\frac{b^2+c^2-a^2}{2bc}-1=\frac{\left(b-c\right)^2-a^2}{2bc}\) ;
\(\frac{c^2+a^2-b^2}{2ac}-1=\frac{\left(c-a\right)^2-b^2}{2ac}\)
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}-1=\frac{\left(a+b\right)^2-c^2}{2ab}+\frac{\left(b-c\right)^2-a^2}{2bc}+\frac{\left(c-a\right)^2-b^2}{2ac}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{2ab}+\frac{\left(b-c-a\right)\left(b-c+a\right)}{2bc}+\frac{\left(c-a-b\right)\left(c-a+b\right)}{2ac}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{2ab}+\frac{\left(b-c-a\right)\left(b-c+a\right)}{2bc}+\frac{\left(a+b-c\right)\left(a-c-b\right)}{2ac}\)
\(=\left(a+b-c\right)\left(\frac{a+b+c}{2ab}+\frac{b-c-a}{2bc}+\frac{a-c-b}{2ac}\right)\)
\(=\left(a+b-c\right)\left[\frac{\left(a+b+c\right)c+\left(b-c-a\right)a+\left(a-c-b\right)b}{2abc}\right]\)
\(=\left(a+b-c\right)\left[\frac{ac+bc+c^2+ab-ac-a^2+ab-bc-b^2}{2abc}\right]\)
\(=\left(a+b-c\right)\left[\frac{c^2-\left(a-b\right)^2}{2abc}\right]\)
\(=\left(a+b-c\right).\frac{\left(c-a+b\right)\left(c+a-b\right)}{2abc}\) ( * )
Vì a ; b ; c là 3 cạnh của tam giác nên biểu thức (*) luôn > 0
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}-1>0\)
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}>1\left(đpcm\right)\) ( 2 )
Từ ( 1 ) ; ( 2 ) => a ; b ; c là 3 cạnh của 1 tam giác