Cho a + b + c =0
CMR: \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)
Cách làm phải thật đơn giản để tớ hiểu nha
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
KB
8 tháng 2 2019
Đây nhé
Đặt b + c = x ; c + a = y ; a + b = z
\(\Rightarrow\hept{\begin{cases}x+y=2c+b+a=2c+z\\y+z=2a+b+c=2a+x\\x+z=2b+a+c=2b+y\end{cases}}\)
\(\Rightarrow\frac{x+y-z}{2}=c;\frac{y+z-x}{2}=a;\frac{x+z-y}{2}=b\)
Thay vào PT đã cho ở đề bài , ta có :
\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\)
\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
( cái này cô - si cho x/y + /x ; x/z + z/x ; y/z + z/y)
BA
17 tháng 11 2019
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
31 tháng 8 2017
Câu hỏi của Bùi Minh Quân - Toán lớp 9 - Học toán với OnlineMath
BT
19 tháng 8 2017
Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{c-a}+\frac{1}{a-b}\)
Tương tự:
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b}{\left(b-c\right)\left(b-a\right)}+\frac{b-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)
Và: \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c}{\left(c-a\right)\left(c-b\right)}+\frac{c-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
=> đpcm
Đặt A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
B = \(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow\)A . B = 9
Ta có : A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
Nhân abc với A ta được:
Aabc = \(\frac{abc\left(a-b\right)}{c}+\frac{abc\left(b-c\right)}{a}+\)\(\frac{abc\left(c-a\right)}{b}\)
Aabc = ab.( a - b ) + bc.( b - c ) + ac.( c - a )
Aabc = ab.( a - b ) + bc.( a - c + b - a ) + ac.( a - c )
Aabc = ab.( a - b ) - bc.( a - b ) - bc.( c - a ) + ac.(c - a )
Aabc = b.( a - b ).( a - c ) - c.( a - b ).(c - a )
Aabc= ( a - b ).( a - c ).( b - c )
A = \(\frac{\left(a-b\right).\left(a-c\right).\left(b-c\right)}{abc}\)
Xét a + b + c = 0 \(\Rightarrow\) a + b = - c ; c + a = -b ; b + c = -a
Nhân ( a - b ).( c - a ).( b - c ) với B ta được :
B( a - b).( c - a ).( b - c ) = \(\frac{c\left(a-b\right).\left(c-a\right).\left(b-c\right)}{a-b}\)+ \(\frac{a\left(a-b\right).\left(b-c\right).\left(c-a\right)}{b-c}\)+ \(\frac{b\left(a-b\right).\left(b-c\right).\left(c-a\right)}{c-a}\)
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) + a.( b - c ).( c - a ) + b.( a - b ).( b - c)
B( a - b ).( c - a ) .( b - c ) = c.( c - a ).( b - c ) + ( a - b ).( -b - c ).( c - a ) + b.( a - b ).( b - c )
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) - b.( a - b ).( c- a ) + b.( a - b ).(b - c ) - c.( a - b ).( c - a )
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( -a + 2b - c ) + b.( a - 2c +b).(a - b )
B( a - b).( c - a ).( b - c ) = -3bc.( b + c - 2a )
B( a - b ).( c - a ).( b - c ) = -9abc
B = \(\frac{9abc}{\left(a-b\right).\left(c-a\right).\left(b-c\right)}\)
NHÂN A VỚI B :
\(\frac{\left(a-b\right).\left(b-c\right).\left(a-c\right)}{abc}\)\(.\)\(\frac{9abc}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)= 9
\(\Rightarrow\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\)\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)
MÌNH CŨNG KHÔNG CHẮC LẮM !