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10 tháng 6 2020

a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

 0,5 x + 0,6 ( x - 2 ) = 3

0,5 x + 0.6 x - 1,2 = 3

1,1 x = 4,2

x = \(\frac{42}{11}\)

Kết luận: 

b) \(\frac{1}{3}x-0,5x=0,75\)

\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(-\frac{1}{6}x=\frac{3}{4}\)

\(x=-\frac{9}{2}\)

Kết luận: 

c) \(\frac{3}{-2}x-0,5x=75\%\)

-1,5x - 0,5x = 0,75 

-2x = 0,75 

x = -0,375

Kết luận: 

d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

-0,4 x + 0,25 = 0,75 - 0,75 x 

-0,4 x + 0,75 x = 0,75 - 0,25

0,35 x = 0,5 

x = \(\frac{10}{7}\)

Kết luận: 

10 tháng 6 2020

\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)

\(\frac{11}{10}x-\frac{6}{5}=3\)

\(\frac{11}{10}x=\frac{21}{5}\)

\(x=\frac{42}{11}\)

\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(\frac{-1}{6}x=\frac{3}{4}\)

\(x=\frac{-9}{2}\)

\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)

\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(-2x=\frac{3}{4}\)

\(x=\frac{-3}{8}\)

\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)

\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)

\(\frac{7}{20}x=\frac{1}{2}\)

\(x=\frac{10}{7}\)

25 tháng 8 2016

a, (2x-1)3=8

2x-1=2

2x=2+1

2x=3

x=1,5

a, \(\left(2x-1\right)^3=8\)

\(\Leftrightarrow\left(2x-1\right)^3=2^3\)

\(\Leftrightarrow2x-1=2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

b, \(\left(x-1\right)^x+2=\left(x-1\right)^x+4\)

\(\Leftrightarrow\left(x-1\right)^x+2-\left(x-1\right)^x-4=0\)

\(\Leftrightarrow-2\ne0\)=> vô nghiệm 

c, \(3^x-1+5.3^{3x}-1=162\)

Đề sai.

8 tháng 3 2022

\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

8 tháng 3 2022

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

                   \(x=\dfrac{-2}{3}\)

vậy x =....

17 tháng 1

Bài 3:

a: \(\frac13+\frac23:x=-7\)

=>\(\frac23:x=-7-\frac13=-\frac{22}{3}\)

=>\(x=\frac23:\frac{-22}{3}=\frac{-2}{22}=-\frac{1}{11}\)

b: \(3\frac12-\frac12x=\frac23\)

=>\(\frac72-\frac{x}{2}=\frac13\)

=>\(7-x=\frac23\)

=>\(x=7-\frac23=\frac{21}{3}-\frac23=\frac{19}{3}\)

c: \(\left\lbrack\left(x+\frac13\right)\cdot\frac34+5\right\rbrack:2=3\)

=>\(\left(x+\frac13\right)\cdot\frac34+5=3\cdot2=6\)

=>\(\left(x+\frac13\right)\cdot\frac34=1\)

=>\(x+\frac13=1:\frac34=\frac43\)

=>\(x=\frac43-\frac13=\frac33=1\)

d: \(\left(1\frac23-x\right)\cdot0,75-2=\frac12\)

=>\(\left(\frac53-x\right)\cdot\frac34=2+\frac12=\frac52\)

=>\(\frac53-x=\frac52:\frac34=\frac52\cdot\frac43=\frac{5\cdot2}{3}=\frac{10}{3}\)

=>\(x=\frac53-\frac{10}{3}=-\frac53\)

e: \(x+75\%\cdot x=-1,6\)

=>\(x+0,75x=-1,6\)

=>1,75x=-1,6

=>\(x=-\frac{160}{175}=-\frac{32}{35}\)

f: \(\left(x-\frac23\right)\left(2x+1\right)=0\)

=>\(\left[\begin{array}{l}x-\frac23=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-\frac12\end{array}\right.\)

g: \(1-\left(2x+\frac12\right)^2=\frac34\)

=>\(\left(2x+\frac12\right)^2=1-\frac34=\frac14\)

=>\(\left[\begin{array}{l}2x+\frac12=\frac12\\ 2x+\frac12=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac12-\frac12=0\\ 2x=-\frac12-\frac12=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\end{array}\right.\)

h: \(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2020}{2021}\)

=>\(1-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2020}{2021}\)

=>\(1-\frac{1}{5x+6}=\frac{2020}{2021}\)

=>\(\frac{1}{5x+6}=\frac{1}{2021}\)

=>5x+6=2021

=>5x=2015

=>x=403