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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{-3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{-3}{ab}\cdot\frac{-1}{c}=\frac{3}{abc}\)
Ta có: \(M=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+3\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{a}\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{a}\frac{1}{b}\left(-\frac{1}{c}\right)\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\frac{1}{abc}=\frac{3}{abc}\)
Ta lại có :
\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{bca}{b^3}+\frac{cab}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
\(\)
Bài làm:
Ta có: \(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
CM HĐT phụ:
Ta có: \(a^3+b^3+c^3=\left(a^3+b^3+c^3-3abc\right)+3abc\)
\(=\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\right]+3abc\)
\(=\left[\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\right]+3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
Áp dụng vào trên ta được:
\(abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\right]\)
Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(P=abc.\frac{3}{abc}=3\)
Vậy P = 3
a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)(1)
Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)nên:
(1) xảy ra\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
ai làm giúp em phép tính này với em làm mãi ko dc ạ
bài 5 tính nhanh
a 100 -99 +98 - 97 + 96 - 95 + ... + 4 -3 +2
b 100 -5 -5 -...-5 ( có 20 chữ số 5 )
c 99- 9 -9 - ... -9 ( có 11 chữ số 9 )
d 2011 + 2011 + 2011 + 2011 -2008 x 4
i 14968+ 9035-968-35
k 72 x 55 + 216 x 15
l 2010 x 125 + 1010 / 126 x 2010 -1010
e 1946 x 131 + 1000 / 132 x 1946 -946
g 45 x 16 -17 / 45 x 15 + 28
h 253 x 75 -161 x 37 + 253 x 25 - 161 x 63 / 100 x 47 -12 x 3,5 - 5,8 : 0,1
Với a,b,c khác 0 và a+b+c=0 ta có
\(A=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}=\frac{ab}{\left(a+b\right)^2-2ab-c^2}+\frac{bc}{\left(b+c\right)^2-2bc-a^2}+\frac{ca}{\left(c+a\right)^2-2ca-b^2}=\frac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\frac{bc}{\left(b+c+a\right)\left(b+c-a\right)-2bc}+\frac{ca}{\left(c+a+b\right)\left(c+a-b\right)-2ca}=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}+-\frac{1}{2}+-\frac{1}{2}=-\frac{3}{2}\)
Vậy A=-3/2
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
a) \(x^3+y^3+z^3-3xyz\)
\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)