\(x^6-1=\left(x^3-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\\ \RightarrowĐPCM\)

\(2005^3+125=\left(2005+5\right)\left(2005^2+2005\cdot5+5^2\right)=2010\left(2005^2+2005\cdot5+5^2\right)⋮2010\)\(x^2+y^2+z^2+3=2\left(x+y+z\right)\\ \Leftrightarrow x^2+y^2+x^2+3=2x+2y+2z\\ \Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\\ \left(x-1\right)^2\ge0;\left(y-1\right)^2\ge0;\left(z-1\right)^2\ge0\\ \Rightarrow\left(x-1\right)^2=\left(y-1\right)^2=\left(z-1\right)^2=0\\ \Rightarrow x-1=y-1=z-1=0\\ \Leftrightarrow x=y=z=1\)

b) \(2005^3+125\)

\(=2005^3+5^3\)

\(=\left(2005+5\right)\left(2005^2-2005.5+5^2\right)\)

\(=2010\left(2005^2-2005.5+5^2\right)\)\(⋮\) 2010

Vậy \(2005^3+125\) chia hết cho 2010

c) \(x^6-1\)

\(=\left(x^3\right)^2-1^2\)

\(=\left(x^3-1\right)\left(x^3+1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\) \(⋮\) \(\left(x-1\right)\) và \(\left(x+1\right)\)

Vậy \(x^6-1\) chia hết cho \(\left(x-1\right)\) và \(\left(x+1\right)\)

Câu 1:

=>\(x^3-2x^2+x-2-3⋮x-2\)

=>\(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

Bài 1:

a: \(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{2}{1-2x}\)

b: Để A>0 thì 1-2x>0

=>2x<1

=>x<1/2

\(\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt{3x+1}-1\right)}{x}=\lim\limits_{x\rightarrow0}\dfrac{6x}{x\left(\sqrt{3x+1}+1\right)}=\lim\limits_{x\rightarrow0}\dfrac{6}{\sqrt{3x+1}+1}=3\)

\(\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x-2\right)}{x+1}=\lim\limits_{x\rightarrow-1}\left(x-2\right)=-3\)

\(\Rightarrow I-J=6\)

Câu 1:

Ta có: \(M\left(x\right)=6x^3+2x^4-x^2+3x^2-2x^3-x^4+1-4x^3\)

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

\(=\left(x^2+1\right)^2\ge1\forall x\)

hay M(x) vô nghiệm(đpcm)

Câu 2:

Ta có: A(0)=5

\(\Leftrightarrow m+n\cdot0+p\cdot0\cdot\left(0-1\right)=5\)

\(\Leftrightarrow m=5\)

Ta có: A(1)=-2

\(\Leftrightarrow m+n\cdot1+p\cdot1\cdot\left(1-1\right)=-2\)

\(\Leftrightarrow5+n=-2\)

hay n=-2-5=-7

Ta có: A(2)=7

\(\Leftrightarrow5+\left(-7\right)\cdot2+p\cdot2\cdot\left(2-1\right)=7\)

\(\Leftrightarrow-9+2p=7\)

\(\Leftrightarrow2p=16\)

hay p=8

Vậy: Đa thức A(x) là 5-7x+8x(x-1)

\(=5-7x+8x^2-8x\)

\(=8x^2-15x+5\)

\(x+\frac{1}{x}=a\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)

\(\Leftrightarrow x^2+2+\frac{1}{x^2}=a^2\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)

=> Đáp án B

Ta có : \(x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=a^2-2\)

Vậy giá trị biểu thức tính theo a là: a^2-2