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ta có \(\left(a+b\right)^2=2\left(a^2+b^2\right)\Rightarrow a^2+b^2+2ab=2\left(a^2+b^2\right)\)
\(\Rightarrow2ab=a^2+b^2\Rightarrow a^2+b^2-2ab=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)(ĐPCM)
Ta có :\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2-a^2-2ab+2b^2-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
\(\Leftrightarrow a=b\)
nhớ tk cho mk nha <:
ta có \(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)
=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
dấu = xảy ra <=> \(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\) (ĐPCM)
\(a^2+b^2+c^2=ab+bc+ca\)
<=> \(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)
=> a-b=0 ; b-c =0 ; a-c=0
=> a=b ; b=c ; c=a
=> a=b=c
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\) (đpcm)
Chứng Minh: a) a2 + b2 >= 2ab với mọi ab
b) x2+2x+3>0 với mọi x
Trình bày rõ ràng giúp tớ nha (toán8)
câu a :
\(a^2+b^{^{ }2}\ge2ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\)
( a - b ) ^ 2 >= 0 là điều hiển nhiên nên suy ra \(a^2+b^2\ge2ab\)với mọi a ,b
câu b :
\(^{x^2+2x+3\ge0\Leftrightarrow x^2+2x+1+2\ge0\Leftrightarrow\left(x+1\right)^2+2\ge0}\)
vì ( x+1 )^2 >= 0 nên (x + 1 )^2 +2 > 0 với mọi x
1) a) \(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(99+98\right)+....\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+.....+2+1\)
\(=\dfrac{100.101}{2}=5050\)
2) a) \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3a^2b+3ab^2-3a^2b+3ab^2=a^3+b^3=VT\)
b) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b+3ab^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)Khi \(a^3+b^3+c^3=3abc\) \(\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
i.i \(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{3}{abc}=3\)iii. \(a^3+b^3+c^3=3abc\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: a=b=c
\(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
TH2: a+b+c=0
\(B=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
a) a3+b3+a2c+b2c-abc
= (a+b)(a2-ab+b2)+c(a2+b2)-abc
=(a+b) [ (a+b)2-3ab]+c.[(a+b)2-2ab]-abc
=(a+b)(a+b)2-3ab(a+b)+c(a+b)2-3abc
=(a+b)2(a+b+c)-3ab(a+b+c)
=(a+b)2.0-3ab.0
=0
b) ax+ay+2x+2y+4
=a(x+y)+2(x+y)+4
=(x+y)(a+2)+4
=(a-2)(a+2)+4
=a2-4+4
=a2
c) A=1+x+x2+...+x49=>Ax=x+x2+x3+...+x50
- A=1+x+x2+...+x49
---> Ax-A=x50-1
d)(a+b)(a+c)+(c+a)(c+b)
=a2+ac+ab+bc+c2+bc+ac+ab
=a2+c2+2ac+2ab+2bc
=2b2+2bc+2ac+2ab
=2b(b+c)+2a(b+c)
=2b(b+c)(b+a)
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\Rightarrow a^2+2ab+b^2=2a^2+2b^2\Rightarrow a^2-2ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a-b=0\Rightarrow a=b\left(đpcm\right)\)
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\\ \Leftrightarrow a^2+b^2+2ab=2a^2+2b^2\\ \Leftrightarrow a^2-2ab+b^2=0\\ \Leftrightarrow\left(a-b\right)^2=0\\ \Leftrightarrow a-b=0\\ \Leftrightarrow a=b\)