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\(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}\)\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2+ab+bc+ca\right)\left(c-a\right)+\left(c^2+ab+bc+ca\right)\left(a-b\right)+\left(a^2+ab+bc+ca\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2c+bc^2+c^2a-ab^2-a^2b-ca^2\right)+\left(c^2a+a^2b+ca^2-bc^2-ab^2-b^2c\right)+\left(a^2b+ab^2+b^2c-ca^2-bc^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a^2b-ca^2\right)+\left(b^2c-bc^2\right)-\left(ab^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(b-c\right)\left(a^2+bc-ab-ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
"Chấm" nhẹ hóng cao nhân ạ :)
P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)
Vì \(a+b+c=0\) \(\Rightarrow\) \(c=-a-b\)
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) , ta có:
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{\left(b^2-bc+ac-a^2\right)}{ab}=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự, \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc};\) \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)
Mặt khác, ta cũng có: từ \(a+b+c=0\), suy ra \(a^3+b^3+c^3=3abc\)
Vậy, \(B=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\) (vì \(a,b,c\ne0\) nên \(abc\ne0\) )
a)Ta có: a3 + b3 + c3 = 3abc
=>a3+b3+c3-3abc=1/2(a+b+c)((a-b)2+(b-c)2+(c-a)2) =0 (dễ dàng phân tích được bạn tự làm)
=>Có 2 trường hợp
a+b+c=0(loại vì a+b+c khác 0 ) hoặc (a-b)2+(b-c)2+(c-a)2 = 0
Mà (a-b)2 , (b-c)2 , (c-a)2 >= 0 với mọi a,b,c
=>để (a-b)2 + (b-c)2 + (c-a)2 = 0
=>a=b=c
Thay trường hợp a=b=c vào P
=> (2017 +1)(2017+1)(2017+1)=20183
b)Tương tự a+b+c=0
=> a3 + b3 + c3 = 3abc
=>\(A=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ac}\)
\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)
\(A=\frac{3abc}{abc}=3\) Do (a3 +b3 + c3=3abc thay vào)
Câu hỏi của Jungkookie - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)
Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)
\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=> \(\frac{ab+bc+ac}{abc}=0\)
=> \(ab+bc+ac=0\)
=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(=\frac{bc}{a^2-ab-ac+bc}+\frac{ca}{b^2-ab-bc+ac}+\frac{ab}{c^2-ac-bc+ab}\)
\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ca}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}-\frac{ca}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca^2-c^2a}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2-ca^2+c^2a+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(c^2a-bc^2\right)-\left(ca^2-b^2c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
b) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-ab-bc+ac}+\frac{c^2}{c^2-bc-ac+ab}\)
\(=\frac{a^2}{a\left(a-b\right)-c\left(a-b\right)}+\frac{b^2}{b\left(b-a\right)-c\left(b-a\right)}+\frac{c^2}{c\left(c-b\right)-a\left(c-b\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2a-b^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)