\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

ngu như con bò tót, ko biết 1+1=2.

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

\(\Leftrightarrow3-2x=-1\)

\(\Leftrightarrow-2x=-1-3\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)

\(\Leftrightarrow200x+24=120x-105\)

\(\Leftrightarrow80x=-129\)

\(\Leftrightarrow x=-\dfrac{129}{80}\)

Vậy \(x=-\dfrac{129}{80}\)

c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)

\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)

\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)

d) \(\left(x+0,2\right)^2+0,75=1\)

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

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

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)

a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)

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

=>\(-\dfrac{1}{2}x=-1\)

=>\(x=-1:(-\dfrac{1}{2})\)

=>\(x=2\)

vậy ...........

b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)

=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)

=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)

=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)

=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)

=>\(x=-\dfrac{192}{80}\)

vậy...................

c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)

=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)

=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)

=>\(\left|x+0,5\right|=\dfrac{11}{20}\)

=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

vậy ....... hoặc.....

d,\((x+0,2)^2+0,75=1\)

=>\(\left(x+0,2\right)^2=1-0,75\)

=>\(\left(x+0,2\right)^2=0,25\)

=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)

vậy..........................

HỌC TỐT NHA !!!!!

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

a: =>1+3x-6=-x+3

=>3x-5=-x+3

=>4x=8

=>x=2(loại)

b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

=>3x-9+2x-4=x-1

=>5x-13=x-1

=>4x=12

=>x=3(loại)

c: =>x^2-2x+4+x^3+8=12

=>x^3+x^2-2x=0

=>x(x^2+x-2)=0

=>x(x+2)(x-1)=0

=>x=0 hoặc x=1

a) Bổ sung cho đầy đủ đề

b) (3x - 1)/4 = (2x - 5)/3

3(3x - 1) = 4(2x - 5)

9x - 3 = 8x - 20

9x - 8x = -20 + 3

x = -17

c) Điều kiện: x ≠ -1/3

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

3.(3x + 1) = -2(x - 3)

9x + 3 = -2x + 6

9x + 2x = 6 - 3

11x = 3

x = 3/11 (nhận)

Vậy x = 3/11