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Bài 2.
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)\(\Rightarrow S_n=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
Bài 1:
\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{14}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}.3+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-2}{5}\)
\(=\dfrac{5}{5}=1\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
2155-(174+2155)+(-68+174)=2155-174-2155-68+174
= -68
( 1 - \(\dfrac{1}{2}\) ) ( 1- \(\dfrac{1}{3}\)) ( 1 - \(\dfrac{1}{4}\)) ( 1 - \(\dfrac{1}{5}\)) = \(\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}.\dfrac{1}{5}\)
= \(\dfrac{1}{120}\)
Mình ps có 2 câu à ^.^!
A = 1.2.3 + 2.3.4 + ....+ 48.49.50
=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)
= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50
= 48.49.50.51
=> A = 48.49.50.51:4 = 12.49.50.51
bài b) làm tương tự nha
1.
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...........+\dfrac{1}{8.9.10}\right)x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+............+\dfrac{2}{8.9.10}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+........+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{8.9}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}.\dfrac{22}{45}\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{11}{45}.x=\dfrac{23}{45}\)
\(\Leftrightarrow x=\dfrac{23}{11}\)
Vậy \(x=\dfrac{23}{11}\) là giá trị cần tìm
2.
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...............+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...........+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{999}{2000}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)
\(\Leftrightarrow x=1999\)
Vậy \(x=1999\) là giá trị cần tìm
\(\Rightarrow\left(1+1+...+1\right)+2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{n\left(n+1\right)}\right)\)[có (n-1) số 1]
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+\left(1-\dfrac{2}{n+1}\right)\)
\(\Rightarrow n-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n\left(n+1\right)}{n+1}-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n^2+n-2}{n+1}\)
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
Hình như đề bài sai rồi ấy!