a )

`VP= (a+b)^3-3ab(a+b)`

     `=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`

     `=a^3+b^3 =VT (đpcm)`

b) 

b) Ta có

a) Ta có:

`VP= (a+b)^3-3ab(a+b)`

     `=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`

     `=a^3 + b^3=VT(dpcm)`

b) Ta có

Bài 3:

a: \(A=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

c: \(C=x^2-2x+y^2-4y+7\)

\(=x^2-2x+1+y^2-4y+4+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và y-2=0

=>x=1 và y=2

Bài 4:

a: \(A=5-8x-x^2\)

\(=-x^2-8x-16+21\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=5-x^2+2x-4y^2-4y\)

\(=-x^2+2x-1-4y^2-4y-1+7\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và 2y+1=0

=>x=1 và y=-1/2

Bài 5:

a: \(a^2+b^2+c^2=ab+ac+bc\)

=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

=>a=b=c

b: \(a^2-2a+b^2+4b+4c^2-4c+6=0\)

=>\(a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

=>a-1=0 và b+2=0 và 2c-1=0

=>a=1 và b=-2 và c=1/2

Bài 1:

a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)

=100+99+98+87+...+2+1

\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)

b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)

\(=2c^2\)

Bài 2:

a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)

\(=a^3+b^3\)

b: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Bài 3:

a: \(A=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

c: \(C=x^2-2x+y^2-4y+7\)

\(=x^2-2x+1+y^2-4y+4+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và y-2=0

=>x=1 và y=2

Bài 4:

a: \(A=5-8x-x^2\)

\(=-x^2-8x-16+21\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=5-x^2+2x-4y^2-4y\)

\(=-x^2+2x-1-4y^2-4y-1+7\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và 2y+1=0

=>x=1 và y=-1/2

Bài 5:

a: \(a^2+b^2+c^2=ab+ac+bc\)

=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

=>a=b=c

b: \(a^2-2a+b^2+4b+4c^2-4c+6=0\)

=>\(a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

=>a-1=0 và b+2=0 và 2c-1=0

=>a=1 và b=-2 và c=1/2

Bài 1:

a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)

=100+99+98+87+...+2+1

\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)

b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)

\(=2c^2\)

Bài 2:

a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)

\(=a^3+b^3\)

b: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Bài 1:

a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)

=100+99+98+87+...+2+1

\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)

b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)

\(=2c^2\)

Bài 2:

a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)

\(=a^3+b^3\)

b: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)

\(2,\\ a,a^3+b^3=a^3=3a^2b+3ab^2+b^3-3a^2b-3ab^2\\ =\left(a+b\right)^3-3ab\left(a+b\right)\\ b,a^3+b^3+c^3-3abc\\ =\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ac-ab-bc\right)\)

khó v. e ko giải đc đâu

Bài 4:

a: \(A=5-8x-x^2\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=5-x^2+2x-4y^2-4y\)

\(=-x^2+2x-1-4y^2-4y-1+7\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và 2y+1=0

=>x=1 và y=-1/2

Bài 3:

a: \(A=4x^2+4x+11\)

\(=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

c: \(C=x^2-2x+y^2-4y+7\)

\(=x^2-2x+1+y^2-4y+4+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và y-2=0

=>x=1 và y=2

+) Ta có: a 3 + b 3 = a + b 3 - 3 a b a + b

Thật vậy, VP = a + b 3  – 3ab (a + b)

= a 3 + 3 a 2 b + 3 a b 2 + b 3 - 3 a 2 b - 3 a b 2

= a 3 + b 3  = VT

Nên  a 3 + b 3 + c 3 = a + b 3 - 3 a b a + b + c 3  (1)

Ta có: a + b + c = 0 ⇒ a + b = - c (2)

Thay (2) vào (1) ta có:

a 3 + b 3 + c 3 = - c 3 - 3 a b - c + c 3 = - c 3 + 3 a b c + c 3 = 3 a b c

Vế trái bằng vế phải nên đẳng thức được chứng minh.

\(\Leftrightarrow a^3+b^3+c^3-3abc>=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc>=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)>=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac>=0\)(vì a+b+c>0)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2>=0\)(luôn đúng)

\(a^3+b^3+c^3\ge3abc\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\ge0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\ge0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)

Vì \(a,b,c>0\Leftrightarrow a+b+c>0\)

Lại có \(a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

Nhân vế theo vế ta được đpcm

Dấu \("="\Leftrightarrow a=b=c\)

a³+b³+c³= (a+b)³-3ab(a+b)+c³
Thay a+b=-c vào, ta được: 
a³ + b³ +c³ = (-c)³ -3ab(-c) +c³ = 3abc (đpcm)

a+b+c=1; a>0; b>0; c>0

=>a>=b>=c>=0

=>a(a-c)>=b(b-c)>=0

=>a(a-b)(a-c)>=b(a-b)(b-c)

=>a(a-b)(a-c)+b(b-a)(b-c)>=0

mà (a-c)(b-c)*c>=0 và c(c-a)(c-b)>=0 

nên a(a-b)(a-c)+b(b-a)(b-c)+(a-c)(b-c)*c>=0

=>a^3+b^3+c^3+3acb>=a^2b+a^2c+b^2c+b^2a+c^2b+c^2a

=>a^3+b^3+c^3+6abc>=(a+b+c)(ab+bc+ac)

=>a^3+b^3+c^3+6abc>=(ab+bc+ac)

mà a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

nên 2(a^3+b^3+c^3)+3acb>=a^2+b^2+c^2>=ab+bc+ac(ĐPCM)