có play

15 cặp

Đặt \(A=1^2+2^2+3^2+...+n^2\)

\(\Rightarrow A=1.1+2.2+3.3+...+n.n\)

\(\Rightarrow A=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+n\left[\left(n+1\right)-1\right]\)

 \(\Rightarrow A=1.2-1+2.3-2+3.4-3+...+n.\left(n+1\right)-n\)

\(\Rightarrow A=\left[1.2+2.3+3.4+...+n.\left(n+1\right)\right]-\left(1+2+3+...+n\right)\)

Đặt \(B=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)

\(\Rightarrow3B=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Đặt \(C=1+2+3+...+n\)

Số số hạng: ( n - 1) : 1 + 1 = n - 1 + 1 = n

\(\Rightarrow C=\frac{n\left(n+1\right)}{2}\)

Thay B và C vào A

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=n.\left(n+1\right).\left[\frac{n+2}{3}-\frac{1}{2}\right]\)

\(\Rightarrow A=n\left(n+1\right).\frac{2\left(n+2\right)-3}{6}=n\left(n+1\right).\frac{2n+4-3}{6}=\frac{n\left(n+1\right).\left(2n+1\right)}{6}\)

Ta có : x - y + x + y = 3x + 1 + 3y

           x + x - y + y = 3x + 3y + 1

           2x - 3x - 3y   = 1

           x - 3y           = 1 = 1 . 1 = ( - 1).( - 1)

\(\Rightarrow\)+ x = 1

            y = 1

         + x = - 1

            y = - 1

Vậy x = 1; y = 1

hoặc x = - 1; y = - 1

b,                                                     Bài giải

\(\left(-32\right)^9=\left(-16\cdot2\right)^9=\left(-16\right)^9\cdot2^9\)

\(\left(-16\right)^{13}=\left(-16\right)^9\cdot\left(-16\right)^4=\left(-16\right)^9\cdot\left[\left(-2\right)^4\right]^4=\left(-16\right)^9\cdot\left(-2\right)^{16}=\left(-16\right)^9\cdot2^{16}\)

Vì \(2^9< 2^{16}\) nên \(\left(-32\right)^9>\left(-16\right)^{13}\)

a) \(\frac{\left(-3\right)^n}{81}=-27\)

     \(\left(-3\right)^n=81.\left(-27\right)\)

     \(\left(-3\right)^n=\left(-3\right)^4.\left(-3\right)^3\)

      \(\left(-3\right)^n=\left(-3\right)^7\)

         \(n=7\)

b) 

\(4< 2^n\le2.16\)

\(2^2< 2^n\le2^5\)

\(2< n\le5\)

\(n\in\left\{3;4;5\right\}\)

c) 

\(\frac{16^n}{4^n}=1024\)

\(\left(\frac{16}{4}\right)^n=4^5\)

\(4^n=4^5\)

\(n=5\)