cho a.b.c đôi một khác nhau cmr:
a.b-c/(a-b)(a-c)+c-a/(b-c)(b-a)+a-b/(c-a)(c-b)=2/a-b+2/b-c+2/c-a
b,(x-b)(x-c)/(a-b)(a-c)+(x-c)(x-a)/(b-c)(b-a)+(x-a)(x-b)/(c-a)(c-b)
K
Khách
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30 tháng 8 2019
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
3 tháng 1 2017
bài 1
ab+bc+ca=0
=>ab+bc=-ca
ta có (a+b)(b+c)(c+a)/abc
=> (ab+ac+bc+b2)(c+a)/abc
=> (0+b2)(c+a)/abc
=>b2c+b2a/abc
=>b(ab+bc)/abc
=>b(-ac)/abc
=>-abc/abc=-1
NN
1 tháng 8 2019
Bài 1:
a ) a.( b2 + c2 ) + b.( a2 + c2 ) + c.( a2 + b2 ) + 2abc
= ab2 + ac2 + a2b + bc2 + a2c + b2c + 2abc
= ( ab2 + a2b ) + ( ac2 + bc2 ) + ( a2c + 2abc + b2c )
= ab.( a + b ) + c2.( a + b ) + c.( a2 + 2ab + b2 )
= ab.( a + b ) + c2.( a + b )v + c.( a + b)2
= ( a + b ).[ ( ab + c2 + c. ( a + b ) ]
= ( a + b ).( ab + c2 + ac + bc )
= ( a + b ).[ ( ab + ac ) + ( c2 + bc) ]
= ( a + b ).[ a.( b + c ) + c.( b + c ) ]
= ( a + b ).( b + c ).( a + c )
b) ab.( a + b ) - bc.( b + c ) + ac.( a - c )
= ab.( a + b ) - bc.( b + c ) + ac.[ ( a + b ) - ( b + c ) ]
= ab.( a + b ) - bc. ( b + c ) + ac.( a + b ) - ac.( b + c )
= ab.( a + b ) + ac.( a + b ) - bc.( b + c ) - ac.( b + c )
= ( a + b ).( ab + ac ) + ( b + c ).( -bc - ac )
= ( a + b ).a.( b + c ) - ( b + c ).c.( a + b )
= ( a + b ).( b + c ).( a - c )
c) ( x2 + x )2 + 2.( x2 + x ) - 3
Đặt x2 + x = a
Khi đó đa thức trở thành:
a2 + 2a - 3
= a2 + 3a - a - 3
= a.( a + 3 ) - ( a + 3 )
= ( a - 1 ).( a - 3 )
\(\Rightarrow\) ( x2 + x - 1 ).( x2 + x - 3 )
B2
ab.( a - b ) + bc.( b - c ) + ca.( c - a ) = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.[ ( a - b ) + ( b - c ) ] = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.( a - b ) - ca.( b - c ) = 0
\(\Leftrightarrow\)ab.( a - b ) - ca.( a - b ) + bc.( b - c ) - ca.( b - c ) = 0
\(\Leftrightarrow\) ( a - b ).( ab - ca ) + ( b - c ).( bc - ca ) = 0
\(\Leftrightarrow\) ( a - b ).a.( b - c ) - ( b - c ).c.( a - b ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) a = b , b = c , a = c
\(\Rightarrow\) a = b = c
13 tháng 7 2019
Ta có: \(\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{z\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{x\left(a-b\right)\left(a-c\right)}-\frac{1}{y\left(a-b\right)\left(b-c\right)}+\frac{1}{z\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{xz\left(a-c\right)}{yxz\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{xy\left(a-b\right)}{zxy\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(a-c\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)\(=\frac{yz\left(b-c\right)-xz\left[\left(b-c\right)+\left(a-b\right)\right]+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(b-c\right)-xz\left(a-b\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(y-x\right)-\left(a-b\right)x\left(z-y\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(c+a-b-b-c+a\right)-\left(a-b\right)x\left(a+b-c-c-a+b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(2a-2b\right)-\left(a-b\right)x\left(2b-2c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)2z\left(a-b\right)-\left(a-b\right)2x\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(2z-2x\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{2\left(z-x\right)}{xyz\left(a-c\right)}=\frac{2\left(a+b-c-b-c+a\right)}{xyz\left(a-c\right)}\)
\(=\frac{2\left(2a-2c\right)}{xyz\left(a-c\right)}=\frac{2.2\left(a-c\right)}{xyz\left(a-c\right)}=\frac{4}{xyz}\Rightarrowđpcm\)
