a) Lm r nkoa!

b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)

\(=\left(0,25x\right):3=\frac{20}{3}\)

\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)

\(\Rightarrow x=20:0,25=80\)

\(\Rightarrow x=80\)

c) \(0,01:2,5=\left(0,75x\right):0,75\)

\(=\frac{1}{250}=\left(0,75x\right):0,75\)

\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)

\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)

\(\Rightarrow x=\frac{1}{250}\)

d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)

\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)

\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)

\(\Rightarrow x=\frac{2}{5}:0,1=4\)

\(\Rightarrow4\)

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

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

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

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

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

=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)

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

=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)

\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\)\(:16\frac{2}{3}=0\)

\(\Rightarrow\)\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0\)

\(\Rightarrow\)\(5\frac{3}{8}+x-7\frac{5}{24}=2\)

\(\Rightarrow5\frac{3}{8}+x=9\frac{5}{24}\)

\(\Rightarrow x=3\frac{5}{6}\)

( Cách làm thì đúng rồi nhưng đáp án tớ ko chắc chắn )

a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)

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

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

\(x=-1\)

b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)

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

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

\(\frac{1}{3}x=\frac{1}{3}\)

\(x=1\)

c) \(2^x+2^{x+1}=24\)

\(2^x+2^x.2=24\)

\(2^x.\left(1+2\right)=24\)

\(2^x.3=24\)

\(2^x=8\)

\(2^x=2^3\)

\(x=3\)

a, (x+1/3)^3 = -8/27

=>(x+1/3)^3 = (-2/3)^3

=>x+1/3     = -2/3

=>x           = -1

b, (1/3x+4/3)^2 = 25/9

=>(1/3x+4/3)^2 = (5/3)^2

=>(1/3x+4/3)   = 5/3

=>1/3x           = 1/3

=>    x           = 1

c, 2^x + 2^x+1 = 24

=>2^x + 2^x . 2 = 24

=>2^x.(1+2)     = 24

=>2^x . 3         = 24

=>2^x              =8

=>2^x             = 2^3

=>  x              = 3

bn tk cho mình nha

a.

\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

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

TH2:

\(x-\frac{3}{4}=0\)

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

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

b.

\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)

TH1:

\(\frac{1}{2}x-3=0\)

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

\(x=3\div\frac{1}{2}\)

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

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

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

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

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

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

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

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

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

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

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

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

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

\(x=5\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{2010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2010}\)
\(\Leftrightarrow x+1=2010\)
\(\Leftrightarrow x=2009\)