Tính giá trị của biểu thức sau . biết a+b+c=0
\(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
2 tháng 12 2016
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b},\)ta có :
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)
\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc},M.\frac{b}{c-a}=1+\frac{2b^3}{abc}.\)
Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)
27 tháng 5 2016
Cách 1 . \(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Đặt \(\frac{a-b}{c}=x\); \(\frac{b-c}{a}=y\) ; \(\frac{c-a}{b}=z\)
Ta có : \(\frac{x+y}{z}=\frac{\frac{a-b}{c}+\frac{b-c}{a}}{\frac{c-a}{b}}=\frac{ab\left(a-b\right)+cb\left(b-c\right)}{ac\left(c-a\right)}=\frac{b\left(b-a-c\right)}{ac}=\frac{2b^2}{ac}=\frac{2b^3}{abc}\)
tương tự : \(\frac{y+z}{x}=\frac{2c^3}{abc}\); \(\frac{x+z}{y}=\frac{2a^3}{abc}\)
\(\Rightarrow A=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)
\(=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Áp dụng bài toán phụ : Nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\) (có thể chứng minh bằng cách rút a = - b - c rồi thay vào tổng ba lập phương) được :
\(A=3+\frac{2}{abc}.3abc=3+6=9\)
LC
27 tháng 5 2016
Đặt \(\frac{a-b}{c}=x=>\frac{c}{a-b}=\frac{1}{x}\)
\(\frac{b-c}{a}=y=>\frac{a}{b-c}=y\)
\(\frac{c-a}{b}=z=>\frac{b}{c-a}=\frac{1}{z}\)
=>\(A=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=x.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=1+\frac{x}{y}+\frac{x}{z}+1+\frac{y}{x}+\frac{y}{z}+1+\frac{z}{x}+\frac{z}{y}\)
=>\(A=3+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)
=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)
Lại có: \(\frac{x+z}{y}=\frac{\frac{a-b}{c}+\frac{c-a}{b}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2}{bc}+\frac{c^2-ac}{bc}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2+c^2-ac}{bc}}{\frac{b-c}{a}}\)
\(=\frac{\frac{\left(ab-ac\right)-\left(b^2-c^2\right)}{bc}}{\frac{b-c}{a}}=\frac{\frac{a.\left(b-c\right)-\left(b+c\right).\left(b-c\right)}{bc}}{\frac{b-c}{a}}=\frac{\frac{\left(a-b-c\right).\left(b-c\right)}{bc}}{\frac{b-c}{a}}\)
\(=\frac{\left(a-b-c\right).\left(b-c\right).a}{\left(b-c\right).bc}=\frac{\left(a-b-c\right).a}{bc}=\frac{\left(a+a-a-b-c\right).a}{bc}\)
\(=\frac{\left[2a-\left(a+b+c\right)\right].a}{bc}\)
Vì a+b+c=0
=>\(\frac{x+z}{y}=\frac{\left(2a-0\right).a}{bc}=\frac{2a^2}{bc}=\frac{2a^3}{abc}\)
Chứng minh tương tự, ta có:
\(\frac{x+y}{z}=\frac{2b^3}{abc}\)
\(\frac{y+z}{x}=\frac{2c^3}{abc}\)
=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{3a^3}{abc}+\frac{3b^3}{abc}+\frac{3c^3}{abc}\)
=>\(A=3+\frac{2a^3+2b^3+2c^3}{abc}\)
=>\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}\)
Vì a+b+c=0
=>a=-(b+c)
=>\(a^3=\left[-\left(b+c\right)\right]^3\)
=>\(a^3=-\left(b+c\right)^3\)
=>\(a^3=-\left[b^3+3bc.\left(b+c\right)+c^3\right]\)
=>\(a^3=-b^3-3bc.\left(b+c\right)-c^3\)
=>\(a^3+b^3+c^3=-3bc.\left(b+c\right)\)
Vì a+b+c=0=>b+c=-a
=>\(a^3+b^3+c^3=-3bc.\left(-a\right)\)
=>\(a^3+b^3+c^3=3abc\)
Thay vào A, ta có:
\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+\frac{6.abc}{abc}=3+6=9\)
=>A=9
Vậy A=9
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 ^^
6 tháng 11 2019
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
Ta có \(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\)
\(=1+\frac{c}{a-b}.\frac{b^2-bc+ca-a^2}{ab}\)
\(=1+\frac{c}{a-b}.\frac{\left(b-a\right)\left(a+b-c\right)}{ab}=1+\frac{2c^2}{ab}\)
Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^2}{bc};M.\frac{b}{c-a}=1+\frac{2b^2}{ca}\)
Do vậy \(A=3+2.\frac{a^3+b^3+c^3}{abc}=9\left(do.a+b+c=0.thi.a^3+b^3+c^3=3abc\right)\)
CW
15 tháng 1 2017
Đặt \(\frac{b+c-a}{c}=\frac{a+b+c}{b}=\frac{b-c+a}{a}=k\)
\(\Rightarrow\hept{\begin{cases}b+c-a=ck\\a+b+c=bk\\b-c+a=ak\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2b=k\left(a+c\right)\left(1\right)\\2c=k\left(b-a\right)\left(2\right)\\2b+2c=b\left(b+c\right)\Rightarrow k=2\end{cases}}\)
Thay k=2 vào (1) và (2) :
\(\hept{\begin{cases}2b=2\left(a+c\right)\\2c=2\left(b-a\right)\end{cases}\Rightarrow\hept{\begin{cases}b=a+c\\c=b-a\Rightarrow a=b-c\end{cases}}}\)
Vậy \(\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{abc}=\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{\left(b-c\right)\left(a+c\right)\left(b-a\right)}=\frac{b+c}{b-c}\)
13 tháng 8 2020
Đặt \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=A\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
<=> \(\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
<=> \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(b-c\right)\left(c-a\right)}+\frac{c}{\left(b-c\right)\left(a-b\right)}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{b}{\left(a-b\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)^2}=0\)
<=> \(A+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(a-b\right)\left(c-a\right)}=0\)
<=> \(A+\frac{\left(a+b\right)\left(a-b\right)+\left(c-a\right)\left(c+a\right)+\left(c+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A+\frac{a^2-b^2+c^2-a^2+b^2-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A=0\)
=> ....
28 tháng 2 2018
đề bài thiếu nhé , a,b,c khác nhau nhé :)
có :\(a=b-c\)
vì a,b,c khác nhau
\(\Rightarrow b-c\ne0\)
có:
\(a+b+c=0\Leftrightarrow c=a-b.\)
\(a=b-c\)
\(b=c-a\)
thày vào M ta được
\(\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)=9\)
N
20 tháng 8 2016
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\)\(\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)\(\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)=\(\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(1)
Tương tự ta có: \(\frac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(2)
\(\frac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(3)
Cộng theo vế của (1);(2)&(3) ta đc:
A\(\le1\)
Dấu''='' xảy ra\(\Leftrightarrow\)a=b=c
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ta có :
\(M\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\)
\(=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)
\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}\)
\(=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc}\)
và \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)
Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)
( Vì \(a^3+b^3+c^3=3abc\). Lại do . ( Phân tích là ra hết ).\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
=> ....
bài này trong sách nâng cao phát triển tập 1