Sửa đề: \(P=ab^2+bc^2+ca^2+abc\le\dfrac{4}{27}\)

Không mất tính tổng quát, giả sử \(a=mid\left\{a;b;c\right\}\)

\(\Rightarrow\left(b-a\right)\left(a-c\right)\ge0\)

\(\Leftrightarrow ab+ac\ge a^2+bc\)

\(\Leftrightarrow ca^2+bc^2\le abc+ac^2\)

Do đó:

\(ab^2+abc+ca^2+bc^2\le ab^2+abc+abc+ac^2=a\left(b+c\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2}.2a\left(b+c\right)\left(b+c\right)\le\dfrac{1}{54}\left(2a+2b+2c\right)^3=\dfrac{4}{27}\)

\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 5:

\(a+b=1\Rightarrow a=1-b\)

\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)

\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 7:

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

\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

giúp mình vs

5.

Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

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

\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)

\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)

6.

Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)

Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)

Mặt khác với mọi a;b ta có:

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)

Từ đó:

\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)

\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)

\(N_{max}=2\) khi \(a=b=1\)

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