x^3 + ax + b x^2 - 2x - 3 x + 2 x^3 - 2x^2 - 3x ax+2x^2 + b + 3x 2x^2 - 4x - 6 ax + 7x + b - 6

Để \(x^3+ax+b⋮x^2-2x-3\)

\(\Rightarrow ax+7x+b-6=0\Leftrightarrow x\left(a+7\right)+\left(b-6\right)=0\)

Đẳng thức xảy ra khi \(a+7=0\Leftrightarrow a=-7;b-6=0\Leftrightarrow b=6\)

a, \(I=s\left(s^2-t\right)+\left(t^2+s\right)=s^3-st+t^2+s\)

Thay t = -1 và s = 1 vào biểu thức trên ta được :

\(1+1+1+1=4\)

b, \(N=u^2\left(u-v\right)-v\left(v^2-u^2\right)=u^2\left(u-v\right)+v\left(u+v\right)\left(u-v\right)\)

\(=\left(u-v\right)\left(u^2+v\left(u+v\right)\right)\)

Thay \(u=0,5=\frac{1}{2};v=-\frac{1}{2}\)

\(=\left(\frac{1}{2}+\frac{1}{2}\right).\frac{1}{4}=\frac{1}{4}\)

a) \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

Tương tự cũng có: \(b^2+c^2\ge2bc,c^2+a^2\ge2ca\).

Cộng lại vế theo vế ta được: 

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Dấu \(=\)khi \(a=b=c\).

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

Dấu \(=\)khi \(a=b\).

c) \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

Dấu \(=\)khi \(a=b\).

Trả lời:

A B D C

a, Vì AB // CD ( gt )

\(\Rightarrow\widehat{A}+\widehat{D}=180^o\)( 2 góc trong cùng phía )

Mà \(\widehat{A}=2\widehat{D}\)( gt )

\(\Rightarrow2\widehat{D}+\widehat{D}=180^o\)

\(\Rightarrow3\widehat{D}=180^o\)

\(\Rightarrow\widehat{D}=60^o\)

\(\Rightarrow\widehat{A}=2\widehat{D}=2\cdot60^o=120^o\)

Vậy \(\widehat{A}=120^0;\widehat{D}=60^o\)

b, Vì AB // CD ( gt )

\(\Rightarrow\widehat{B}+\widehat{C}=180^o\)( 2 góc trong cung phía )

Mà \(\widehat{B}-\widehat{C}=30^o\)\(\Rightarrow\widehat{B}=30^o+\widehat{C}\)

\(\Rightarrow30^o+\widehat{C}+\widehat{C}=180^o\)

\(\Rightarrow2\widehat{C}=150^o\)

\(\Rightarrow\widehat{C}=75^o\)

\(\Rightarrow\widehat{B}=30^o+75^o=105^o\)

Vậy \(\widehat{B}=105^o;\widehat{C}=75^o\)

c, Vì AB // CD ( gt )

\(\Rightarrow\widehat{A}+\widehat{D}=180^o\)( 2 góc trong cùng phía )

Mà \(\widehat{A}-\widehat{D}=10^o\)( gt ) \(\Rightarrow\widehat{A}=10^o+\widehat{D}\)

\(\Rightarrow10^o+\widehat{D}+\widehat{D}=180^o\)

\(\Rightarrow2\widehat{D}=170^o\)

\(\Rightarrow\widehat{D}=85^o\)

\(\Rightarrow\widehat{A}=10^o+85^o=95^o\)

Vì AB // CD ( gt )

\(\Rightarrow\widehat{B}+\widehat{C}=180^o\)( 2 góc trong cung phía )

Mà \(\widehat{B}=3\widehat{C}\)( gt )

\(\Rightarrow3\widehat{C}+\widehat{C}=180^o\)

\(\Rightarrow4\widehat{C}=180^o\)

\(\Rightarrow\widehat{C}=45^o\)

\(\Rightarrow\widehat{B}=3\widehat{C}=3\cdot45^o=135^o\)

Vậy \(\widehat{A}=95^o;\widehat{B}=135^o;\widehat{C}=45^o;\widehat{D}=85^o\)

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

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}=\frac{2}{ab}\)

Đặt \(x=\frac{1}{a},y=\frac{1}{b}\)

\(x^3+y^3=2xy\)

\(\Leftrightarrow x^4+xy^3=2x^2y\)

\(\Leftrightarrow x^4-2x^2y+y^2=y^2-xy^3\)

\(\Leftrightarrow\left(x^2-y\right)^2=y^2\left(1-xy\right)\)

\(\Leftrightarrow1-xy=\left(\frac{x^2-y}{y}\right)^2\)

Từ đó ta có đpcm. 

Ta có:\(a^3+b^3=2a^2b^2\)

Đặt \(\frac{a}{b^2}=x;\frac{b}{a^2}=y\left(x,y\in Q\right)\)

Ta được:\(\hept{\begin{cases}x+y=2\\xy=\frac{1}{ab}\end{cases}}\)

Theo Viet đảo ,x và y là nghiệm:

\(t^2-2t+\frac{1}{ab}=0\)

\(\Delta'=1-\frac{1}{ab}\)

Do \(x;y\)là số hữu tỉ 

\(\Rightarrow\sqrt{\Delta'}\)hữu tỉ\(=\sqrt{1-\frac{1}{ab}}\)hữu tỉ

Cre:h

\(a^3-8ab-16b^3=0\)

\(\Leftrightarrow a^4-8a^2b=16ab^3\)

\(\Leftrightarrow a^4-8a^2b+16b^2=16ab^3+16b^2\)

\(\Leftrightarrow\left(a^2-4b\right)^2=16b^2\left(ab+1\right)\)

\(\Leftrightarrow1+ab=\left(\frac{a^2-4b}{4b}\right)^2\)

Do đó ta có đpcm.

a) \(3x^2+2y⋮11\Leftrightarrow16\left(3x^2+2y\right)⋮11\Leftrightarrow48x^2-33x^2+32y-44y⋮11\)

\(\Leftrightarrow15x^2-12y⋮11\)

b) \(2x+3y^2⋮7\Leftrightarrow10\left(2x+3y^2\right)⋮7\Leftrightarrow20x-14x+30y^2-14y^2⋮7\)

\(\Leftrightarrow6x+16y^2⋮7\)

\(a>a^2\)khi a là số nguyên âm

\(a< a^2\)khi a là số nguyên dương