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\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
Bài 1: a) Ta có : \(\dfrac{-3}{x}=\dfrac{x}{-27}\Leftrightarrow\left(-3\right).\left(-27\right)=x.x\Leftrightarrow81=x^2\Leftrightarrow9^2=x^2\Leftrightarrow x=9\)
b) Do \(\dfrac{2}{3}\) của x là -150 nên x là: (-150) : \(\dfrac{2}{3}\) = -225
c) \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{18}\)
\(\Leftrightarrow x+2=18\)
\(\Leftrightarrow x=16\)
Bài 2:
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right).0\)
\(A=0\)
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
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!!!
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) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
Bài 3:
Để A là số nguyên thì \(n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)