1) Giải:

a) \(\left(x+2\right)^2-2\left(x+2\right)\left(x-8\right)+\left(x-8\right)^2\)

= \(\left(x+2-x+8\right)^2=10^2=100\)

Vậy biểu thức trên không phụ thuộc vào biến.

b) \(\left(x+y-z-t\right)^2-\left(z+t-x-y\right)^2\)

= \(\left(x+y-z-t-z-t+x+y\right)\left(x+y-z-t+z+t-x-y\right)\)

= \(\left(2x+2y-2z-2t\right).0=0\)

Vậy biểu thức trên không phụ thuộc vào biến.

2) Giải:

Ta có: \(n^3-n=n\left(n^2-1\right)=n.\left(n-1\right).\left(n+1\right)\)

Tích của 3 số nguyên liên tiếp sẽ chia hết cho 6.

Mà n-1 ; n và n+1 là ba số nguyên liên tiếp ( n \(\in\) Z )

Nên n(n-1)(n+1) chia hết cho 6 hay n³-n chia hết cho 6.

3) Giải:

Ta có: \(x+3y=xy+3\)

\(\Leftrightarrow x+3y-xy-3=0\)

\(\Leftrightarrow-x\left(y-1\right)+3\left(y-1\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=0\\3-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=3\end{matrix}\right.\)

Vậy ( x ; y ) = ( 3 : 1 )

C = \(\sqrt{\sqrt{2}+2\sqrt{\sqrt{2}-1}}+\sqrt{\sqrt{2}-2\sqrt{\sqrt{2}-1}}\)

C = \(\sqrt{\sqrt{2}-1+2\sqrt{\sqrt{2}-1}+1}+\sqrt{\sqrt{2}-1-2\sqrt{\sqrt{2}-1}+1}\)

C = \(\sqrt{\left(\sqrt{2}-1+1\right)^2}+\sqrt{\left(\sqrt{2}-1-1\right)^2}\)

C = \(\sqrt{2}-1+1-\sqrt{2}+1+1\)

C = \(2\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\) (1)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{c}=\dfrac{2b}{2d}=\dfrac{a-2b}{c-2d}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a+b}{a+d}=\dfrac{a-2b}{c-2d}\)

\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2009}\)

\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)

\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2009\)

\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2008\)

\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2008\)

\(\Leftrightarrow A^2=2009\)

\(\Leftrightarrow A=\sqrt{2009}\) khi x, y > 0 hoặc \(A=-\sqrt{2009}\) khi x, y < 0

\(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)

\(x^2y^2+\left(x^2+1\right)\left(y^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)

\(x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2018\)

\(x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=2017\)

\(\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)

\(\Rightarrow A=\left(x\sqrt{y^2+1}+y\sqrt{x^2+1}\right)^2=2017\)

\(\Rightarrow A=\sqrt{2017}\) khi x, y > 0 hoặc \(A=-\sqrt{2017}\) khi x, y < 0

a1. A = \(1+4+4^2+4^3+...+4^{58}+4^{59}\)

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

A = \(5+4^2.5+...+4^{58}.5\)

A = \(5\left(1+4^2+...+4^{58}\right)⋮5\)

a2. A = \(1+4+4^2+4^3+...+4^{58}+4^{59}\)

A = \(\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{57}+4^{58}+4^{59}\right)\)

A = \(\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{57}\left(1+4+4^2\right)\)

A = \(\left(1+4+4^2\right)\left(1+4^3+...+4^{57}\right)\)

A = \(21.\left(1+4^3+...+4^{57}\right)⋮21\)

a3. A = \(1+4+4^2+4^3+...+4^{58}+4^{59}\)

A = \(\left(1+4+4^2+4^3\right)+\left(4^4+4^5+4^6+4^7\right)+...+\left(4^{56}+4^{57}+4^{58}+4^{59}\right)\)

A = \(\left(1+4+4^2+4^3\right)+4^4\left(1+4+4^2+4^3\right)+...+4^{56}\left(1+4+4^2+4^3\right)\)

A = \(\left(1+4+4^2+4^3\right)\left(1+4^4+...+4^{56}\right)\)

A = \(85.\left(1+4^4+...+4^{56}\right)⋮85\)

Câu B sao thứ tự số mũ chẳng có quy luật vậy, sao mà làm được :v

minh cung dng muon giai bai nay ai giup voi 

minh tck cko