ta có 

\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)

ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)

\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)

hay \(S_3=0+0+..+0=0\)

\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)

\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)

\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)

\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)

Tính P = Chúc bạn học tốt nhé

P=1+1/3+1/6+1/10+…..+1/2021×2022÷2

P/2=1/2+1/6+1/12+1/20+…..+1/2021×2022

P/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022

P/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022

P=2021/1011

Chúc bn học tốt

Ta có:

\(C=\dfrac{2n-3}{n-2}=\dfrac{2n-4+1}{n-2}=2+\dfrac{1}{n-2}\)

\(C\in Z\Leftrightarrow\dfrac{1}{n-2}\in Z\Leftrightarrow n-2\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow...\)

a) \(\left(-1\right)\left[5^2-\left(4^3\right)\right]=\left(-1\right)\left(25-64\right)=\left(-1\right)\left(-39\right)=39\)

b) \(71.64-32.\left(-7\right)+32.11=142.32+32.7+32.11\)

\(=32.\left(142+7+11\right)=32.\left(142+18\right)=32.160=5120\)

c) \(666-\left(-422\right)-100-88=666+422-100-88\)

\(=\left(666+422\right)-\left(100+88\right)=1088-188=900\)

d) \(23-501-343+61-257+16-499\)

\(=\left(23+61+16\right)-\left(501+499\right)-\left(343+257\right)\)

\(=100-1000-600=100-\left(1000+600\right)=100-1600=-1500\)

\(3^x+4\cdot3^{x-2}=333\)

\(\Rightarrow3^{x-2+2}+4\cdot3^{x-2}=333\)

\(\Rightarrow3^{x-2}\cdot\left(3^2+4\right)=333\)

\(\Rightarrow3^{x+2}\cdot\left(9+4\right)=333\)

\(\Rightarrow3^{x+2}\cdot13=333\)

\(\Rightarrow3^{x+2}=333:13\)

\(\Rightarrow3^{x+2}=\dfrac{333}{13}\)

Không có x nào thỏa mãn

⇒ x ∈ ∅

Mk cần gấp á. Ai hỗ trợ mk với

\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)

\(=\dfrac{75}{101}\)

\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)

\(=\dfrac{100}{2}=50\)

Tính nhanh:

a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{3}{4}\times\dfrac{100}{101}\)

= \(\dfrac{75}{101}\)

b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)

\(=\dfrac{100}{2}\)

\(=50\)

4⁵ x 9⁴ - 2 x 6⁹=6718464 - 20155392 = -13436928

2¹⁰ x 3⁸ + 6⁸ x 20=6718464+33592320=40310784

\(\left(x-36\right):\left(2.3^2\right)=2^2.3\)

\(\Rightarrow\left(x-36\right):18=12\)

\(\Rightarrow\left(x-36\right)=12.18\)

\(\Rightarrow x-36=216\)

\(\Rightarrow x=216+36\)

\(\Rightarrow x=252\)

(x-36):(2.3²)=2.2³

⇒ (x-36):(2.9)=2.8

⇒ (x-36):18=16

⇒ x-36=18.16=288

⇒ x=288+36=324

\(\left(x-36\right):\left(2\cdot3^2\right)=2^2\cdot3\\ x-36=2^2\cdot3\cdot2\cdot3^2\\ x-36=3^3\cdot2^3\\ x-36=216\\ x=216+36\\ x=252\)