Bài 2 :

A = 12 + 14 + 16 + x \(⋮\) 2

mà 12 \(⋮\) 2

14 \(⋮\) 2

16 \(⋮\) 2

\(\Rightarrow\) ( 12 + 14 + 16 ) \(⋮\) 2

\(\Rightarrow\) x \(⋮\) 2

x = 2k ( k \(\in\) N )

A = 12 + 14 + 16 + x \(⋮̸\) 2

mà 12 \(⋮\) 2

14 \(⋮\) 2

16 \(⋮\) 2

\(\Rightarrow\) x \(⋮̸\) 2

x = 2k + r ( k \(\in\) N , r \(\in\) N* )

Bài 3 : Cách làm tương tự như bài 2

1, Ta có: \(\left(n-2\right)^2=\left(n-2\right)^4\)

\(\Rightarrow\left(n-2\right)^2-\left(n-1\right)^4=0\)

\(\Rightarrow\left(n-2\right)-\left(n-2\right)=0\)

\(\Rightarrow n=2\)

Vậy ...

2)

Ta có: 20 \(⋮\) 2n+1

\(\Rightarrow\) 2n+1 \(\in\) Ư(20)

\(\Rightarrow\) 2n+1 \(\in\) {1; 2; 4; 5; 10; 20}

\(\Rightarrow\) 2n \(\in\) {0; 1; 3; 4; 9; 19}

\(\Rightarrow\) n= 2

Lời giải:

Theo định lý Be-du thì số dư của \(P(x)=ax^3+bx^2+c\) khi chia cho \(x+2\) là:

\(P(-2)=-8a+4b+c=0\) (1)

Gọi đa thức thương khi chia $P(x)$ cho\(x^2-1\) là \(Q(x)\). Khi đó ta có:

\(ax^3+bx^2+c=(x^2-1)Q(x)+x+5\)

Thay \(x=\pm 1\) ta thu được:

\(\left\{\begin{matrix} a+b+c=0.Q(1)+6=6(2)\\ -a+b+c=0.Q(-1)+4=4(3)\end{matrix}\right.\)

Từ \((1)(2)(3)\Rightarrow \left\{\begin{matrix} a=1\\ b=1\\ c=4\end{matrix}\right.\)

Vậy \((a,b,c)=(1,1,4)\)

Câu 1: x= -4 ; 4

Câu 2: n= 991

Câu 1: x = + 4

Câu 2: n = 991

sao không ai trả lời vậy ?

Bài 3: 

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(A=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

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

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

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 \(\left(x+1\right)\)

Bài 1:

\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

\(< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1+1-\dfrac{1}{50}\)

\(=2-\dfrac{1}{50}\)

\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)

\(\Rightarrow A< 2\left(đpcm\right)\)

Vậy...