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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) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)
\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)
\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)
a, \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{3}{5}+0,415\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{203}{200}\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{203}{200}+8\)
\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{1803}{200}\)
\(\left(\dfrac{13}{4}-30,75\right).x=\dfrac{1803}{200}\)
\(\dfrac{-55}{2}.x=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}\)
\(x=\dfrac{-1803}{5500}\)
Nếu là tìm số nguyên thì hình như đề sai rồi bạn
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b, \(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
Cho \(A=4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(A=4\dfrac{1}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{13}{3}.\dfrac{-1}{3}\)
\(A=\dfrac{-13}{9}\)
Cho \(B=\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\left(\dfrac{-1}{6}-\dfrac{3}{4}\right)\)
\(B=\dfrac{2}{3}.\dfrac{-11}{12}\)
\(B=\dfrac{-11}{18}\)
Ta có: \(A\le x\le B\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
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 ^.^!
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!!!
Bài 1:
a: \(A=\dfrac{\left(85+\dfrac{7}{30}-83-\dfrac{5}{18}\right):\dfrac{8}{3}}{\dfrac{1}{25}}\)
\(=\left(2+\dfrac{7}{30}-\dfrac{5}{18}\right)\cdot\dfrac{3}{8}\cdot25\)
\(=\dfrac{180+21-25}{90}\cdot\dfrac{75}{8}\)
\(=\dfrac{176}{90}\cdot\dfrac{75}{8}=\dfrac{55}{3}\)
=>12,5% của A là 55/8x1/8=55/64
b: \(B=\dfrac{\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{36}{5}}{19.75:2.5}\)
\(=\dfrac{\left(3+\dfrac{27}{70}\right)\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{\dfrac{210+27}{70}\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{4266}{175}\cdot\dfrac{10}{79}=\dfrac{108}{35}\)
=>5% là 108/35x1/20=27/175
Ta có :
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.................+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+............+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+............+\dfrac{2}{2\left(n+1\right)}=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2003}{4008}\)
\(\dfrac{1}{n+1}=\dfrac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4007\) (Thỏa mãn \(n\in N\))
Vậy \(n=4007\) là giá trị cần tìm
~~Chúc bn học tốt~~
hình như sai đề phải bạn ạ