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BM
9 tháng 7 2019
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
9 tháng 7 2019
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
VM
5 tháng 9 2017
Ta có:
(-3/2:3/-4)*(-9/2)-1/4<x/8<-1/2:3/4:1/8+1
Xét VT = (-3/2.-4/3).(-9/2)-1/4
= 2.-9/2-1/4
=-9-1/4=-37/4=--222/24
Xét VP = -1/2:3/4:1/8+1
=-1/2.4/3.8+1
=-16/3+1
=-13/3=-104/24
=>-222/24<x/8<-104/24=>-222/24<x.3/24<-104/24=>-222<x.3<-104
=>x.3={-221;-220;...;--105}Mà x.3 chia hết cho 3=>x.3 thuộc{-219;-216;...;-105}
=>x={-73;-72;.....-35}
Vậy ..........
31 tháng 8 2022
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8
2 tháng 7 2020
Bài 1:
a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)
hay \(x=\frac{77}{72}\)
Vậy: \(x=\frac{77}{72}\)
b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)
\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)
hay \(x=-\frac{46}{35}\)
Vậy: \(x=-\frac{46}{35}\)
c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)
\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)
hay \(x=\frac{2}{3}\)
Vậy: \(x=\frac{2}{3}\)
d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)
\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)
hay \(x=\frac{25}{42}\)
Vậy: \(x=\frac{25}{42}\)
e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)
\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)
\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)
hay \(x=-\frac{4}{5}\)
Vậy: \(x=-\frac{4}{5}\)
f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)
\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)
\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)
Vậy: \(x=-\frac{2}{3}\)
g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)
\(1\frac{3}{8}-x=\frac{1}{3}\)
\(\frac{11}{8}-x=\frac{1}{3}\)
\(x=\frac{11}{8}-\frac{1}{3}\)
\(x=\frac{25}{24}\)
\(\frac{11}{8}\)-x=\(\frac{1}{3}\)
x=\(\frac{11}{8}\)-\(\frac{1}{3}\)
x=\(\frac{25}{24}\)
tk nhé