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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
\(2005^3-1=\left(2005-1\right)\left(2005^2+2005+1\right)=2004\times\left(2005^2+2005+1\right)⋮2004\left(\text{đ}pcm\right)\)
\(2005^3+125=\left(2005+5\right)\left(2005^2-2005\times5+5^2\right)=2010\times\left(2005^2-2005\times5+5^2\right)⋮2010\)
\(x^6+1=\left(x^2+1\right)\left(x^4-x^2+1\right)⋮x^2+1\left(\text{đ}pcm\right)\)
\(x^6-y^6=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)⋮x-y;x+y\left(\text{đ}pcm\right)\)
a) \(\left(x-y\right)\left(x+y\right)\)
\(=x^2+xy-xy-y^2\)
\(=x^2-y^2\)
b) \(\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)\)
\(=x^4+x^2y^2+x^3y+xy^3-x^3y-xy^3-x^2y^2-y^4\)
\(=x^4-y^4\)
c)\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)
\(=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2-abc\)
\(=2abc+a^2b+a^2c+ab^2+b^2c+bc^2+ac^2\left(1\right)\)
\(\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=a^2+ac+ab+bc\left(b+c\right)\)
\(=a^2b+abc+ab^2+b^2c+a^2c+ac^2+abc+bc^2\)
\(=2abc+a^2b+ab^2+b^2c+a^2c+ac^2+bc^2\left(2\right)\)
Từ (1)(2) => đpcm
đẽ thu gọn vế vd a) ta có vt: ( x-y) .(x+y)=x^2 -y^2
=vp
->dpcm
b) (x-y) . (x^3 +xy^2 +x^2y+y^3)
=(x-y ).(x^3 + y^3)
= x.x^3 -y.y^3
=x^4 - y^4 =vp
->dpcm
c) (a +b+ c) (ab +bc +ac) -abc
=nhân vô rút gọn
=(a^2b +2abc +c^b) +(a^2c+c^2a) + (ab^2+b^2c )
=b(a+c)^2 +ac(a+c) +b^2 (a+c)
=(a+c).[b(a+c)+b^2 +ac+b^2]
=(a+c)(ab+b^2+bc+ac)
=(a+c) [b(a+b)+c(a+b)]
=(a+b)(a+c)(b+c)=vp
->dpcm
4)
a) Ta có \(2^{10}+2^{11}+2^{12}\)
\(=2^{10}\left(1+2+4\right)=2^{10}\cdot7⋮7\)
Vậy: \(2^{10}+2^{11}+2^{12}\) chia hết cho 7(đpcm)
b) Ta có: 7*32=224=25+26+27
3/ \(x^5+y^5\ge x^4y+xy^4\)
\(\Leftrightarrow x^4\left(x-y\right)-y^4\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(x^4-y^4\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
bài 1
theo bài ra ta có
a + b + c = 0 => c = -[a+b] [ 1 ]
Thay (1) vao a^3+b^3+c^3 ta có:
a^3+b^3+[-(a+b)]^3=3ab[-(a+b)]
<=>a^3+b^3-(a+b)=-3ab(a+b)
<=> a3+ b3- a3 -3a2b- 3ab2- b3= -3a2b- 3ab2
<=> 0= 0
vậy ta có đpcm.
\(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Min A = 10 khi: 2x + 1 = 0
<=> x = -1/2