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1) a) a^2+b^2=ab+ba
<=> a^2+b^2-2ab=0
<=> (a-b)^2=0
<=> a-b=0 <=> a=b (đpcm)
b) a^2+b^2+c^2=ab+bc+ca
<=> 2a^2+2b^2+2c^2=2ab+2bc+2ca
<=> (a^2-2ab+b^2)+(a^2-2ca+c^2)+(b^2-2bc+c^2)=0
<=> (a-b)^2+(a-c)^2+(b-c)^2=0
<=> a-b=0 và a-c=0 và b-c=0
<=> a=b và a=c và b=c
<=> a=b=c (đpcm)
(a - b)2 + (b - c)2 + (c - a)2 = 3(a2 + b2 + c2 - ab - bc - ca)
<=> (a - b)2 + (b - c)2 + (c - a)2 = \(\dfrac{3}{2}\)(2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca)
<=> (a - b)2 + (b - c)2 + (c - a)2 = \(\dfrac{3}{2}\)[(a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2)]
<=> (a - b)2 + (b - c)2 + (c - a)2 = \(\dfrac{3}{2}\)[(a - b)2 + (b - c)2 + (c - a)2]
<=> \(\dfrac{1}{2}\)[(a - b)2 + (b - c)2 + (c - a)2] = 0
<=> a = b = c
Cách 2 :
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=3\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=3\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=3\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a;b;c\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\end{matrix}\right.\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
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) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
Ta có : a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
=> (a - b)2 + (b - c)2 + (c - a)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
b) Ta có : 2(x2 + t2) + (y + t)(y - t) = 2x(y + t)
=> 2x2 + 2t2 + y2 - t2 = 2xy + 2t
=> 2x2 + t2 + y2 = 2xt + 2xy
=> 2x2 + t2 + y2 - 2xt - 2xy = 0
=> (x2 - 2xy + y2) + (x2 + t2 - 2xt) = 0
=> (x - y)2 + (x - t)2 = 0
=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)
c) Ta có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = 0
=> a = b = c = 0
Khi đó A = (0 - 1)2003 + 02004 + (0 + 1)2005
= - 1 + 0 + 1 = 0
Vậy A = 0
\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)
\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)
a2+b2+c2=ab+ac+bc
<=>2a2+2b2+2c2=2ab+2ac+2bc
<=>a2-2ab+b2+a2-2ac+c2+b2-2bc=0
<=>(a-b)2+(a-c)2+(b-c)2=0
<=>a-b=0 và a-c=0 và b-c=0
<=>a=b=c
Bài 2:
\(A=\left(2ac-a^2-c^2+b^2\right)\left(2ac+a^2+c^2-b^2\right)\)
\(=\left[b^2-\left(a-c\right)^2\right]\left[\left(a+c\right)^2-b^2\right]\)
\(=\left(b-a+c\right)\left(b+a-c\right)\left(a+c-b\right)\left(a+c+b\right)\)>0
a+b+c=0
=>(a+b+c)3=0
=>a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+6abc=0
=>a3+b3+c3+(3a2b+3ab2+3abc)+(3b2c+3bc2+3abc)+(3a2c+3ac2+3abc)-3abc=0
=>a3+b3+c3+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a3+b3+c3=3abc(ĐPCM)
a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\rightarrow a=b=c=1\)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\rightarrow a=b=c\)