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a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
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) 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) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
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) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)
\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)
\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)
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: \(A\le x\le B\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
a/ Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...........+\dfrac{1}{n^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)
\(\Leftrightarrow A< 1\)
b/ Ta có :
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.................+\dfrac{1}{\left(2n\right)^2}\)
\(=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\right)\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{\left(n-1\right)n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
\(\Leftrightarrow B< \dfrac{1}{2}\)
\(\)\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(A< 1-\dfrac{1}{n}< 1\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2n^2}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B=\dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B< \dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{\left(n-1\right)n}\right)\)
Câu 1:
a, \(\left|-5\right|=5\)
b, \(\left|10\right|=10\)
c, \(\left|-5\right|-\left|10\right|=5-10=-5\)
d, -15.30= -450
Câu 2:
a, Ta có: \(\dfrac{10}{21}.\dfrac{14}{25}=\dfrac{10.14}{21.25}=\dfrac{5.2.7.2}{3.7.5.5}=\dfrac{2.2}{3.5}=\dfrac{4}{15}\)
c, Ta có: \(-\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{-5.2+3.3}{12}=\dfrac{-10+9}{12}=\dfrac{-1}{12}\)
d, \(\dfrac{11}{17}.\dfrac{3}{2017}+\dfrac{11}{17}.\dfrac{2014}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}\left(\dfrac{3}{2017}+\dfrac{2014}{2017}\right)-1\dfrac{11}{17}\)
\(=\dfrac{11}{17}.\dfrac{2017}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}-1-\dfrac{11}{17}=-1\)
Câu 7: a, Để A có nghĩa khi \(x+2\ne0\) \(\Leftrightarrow x=-2\)
b, Ta có: \(A=2\)
<=> \(\dfrac{x-1}{x+2}=2\)
<=> \(\dfrac{x-1}{x+2}-2=0\)
<=> \(\dfrac{x-1}{x+2}-\dfrac{2x+4}{x+2}=0\)
<=> \(\dfrac{x-1-2x-4}{x+2}=0\)
<=> \(\dfrac{-x-5}{x+2}=0\)
<=> -x-5=0
<=> -x=5
<=> x= -5
Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm
\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{13}{3}.\dfrac{-1}{3}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)
\(\dfrac{-13}{9}\)\(\le x\le\)\(\dfrac{-11}{18}\)
\(\dfrac{-26}{18}\)\(\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x\in\left\{\dfrac{-12}{18};\dfrac{-13}{18};\dfrac{-14}{18};\dfrac{-15}{18};...;\dfrac{-24}{18};\dfrac{-25}{18}\right\}\)Tick hộ mình nha bạn
Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$2[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2(\frac{1}{2}-\frac{1}{n+1})=\frac{2022}{2023}$
$1-\frac{2}{n+1}=1-\frac{1}{2023}$
$\Rightarrow \frac{2}{n+1}=\frac{1}{2023}$
$\Rightarrow n+1=2.2023=4046$
$\Rightarrow n=4045$