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a. \(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)( đpcm )
Bài 1:
a)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
b)\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(\Rightarrow-2a^2-2b^2-2c^2+2ab+2bc+2ca=0\)
\(\Rightarrow-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\le0\)
Khi \(a=b=c\)
c)\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
Bài 2:
Từ \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow-2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Rightarrow ab+bc+ca=-1\)\(\Rightarrow\left(ab+bc+ca\right)^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+b^2ca+c^2ab\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=1\left(vi`....a+b+c=0\right)\)
Khi đó: \(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)
\(\Rightarrow a^4+b^4+c^4+2=4\Rightarrow a^4+b^4+c^4=2\)
so u cn tk m sl fr u
a2 + b2+ c2 = ab + bc + ca
=> a2 + b2+ c2 -ab - bc - ca = 0
=> 2 ( a2 + b2 + c2 -ab -bc - ca) =0
=> ( a2 - 2ab + b2 ) + ( b2 -2bc + c2 ) + ( c2 - 2ca + a2 ) = 0
<=> ( a-b )2 + ( b -c)2 + ( c- a)2 =0
Do ( a -b)2 \(\ge\)0 ( b-c)2 + \(\ge\)0 ( c -a )2 \(\ge\)0
=> a-b =0 ; b -c = 0 ; c -a = 0
=> a=b ; b = c ; c =a
Vậy a = b = c
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