Giải:

\(0,28-0,3:\left(50\%x-1\dfrac{1}{3}\right)=-1\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{7}{25}-\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=-\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=\dfrac{7}{25}-\left(-\dfrac{5}{3}\right)\)

\(\Leftrightarrow\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=\dfrac{146}{75}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{4}{3}=\dfrac{3}{10}:\dfrac{146}{75}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{4}{3}=\dfrac{45}{292}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{45}{292}-\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{1}{2}x=-\dfrac{1033}{876}\)

\(\Leftrightarrow x=-\dfrac{1033}{876}:\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1033}{438}\)

Vậy \(x=-\dfrac{1033}{438}\).

Chúc bạn học tốt!

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{2}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Giải:

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\\\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=\dfrac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{21}{10}\\x=\dfrac{9}{10}\end{matrix}\right. \)

Vậy \(x=\dfrac{21}{10}\) hoặc \(x=\dfrac{9}{10}\)

1/2+1/3<x<=1+1/2+1/5

=>5/6<x<=1+7/10

=>5/6<x<17/10

mà x là số nguyên

nên x=1

a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)

\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)

b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)

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

\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)

c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)

\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)

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

a: (1/2x^2+y^2)^2

=(1/2x^2)^2+2*1/2x^2*y^2+y^4

=1/4x^4+x^2y^2+y^4

b: (4/5x^2-2/3y)^2

=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2

=16/25x^4-16/15x^2y+4/9y^2

c: =(2x)^2-(1/2)^2

=4x^2-1/4

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)

\(=x^3+2^3\)

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

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

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125

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

=> x +\(\dfrac{1}{4}\)= \(\dfrac{14}{15}\)

=> x = \(\dfrac{14}{15}\) - \(\dfrac{1}{4}\)

=> x = \(\dfrac{41}{60}\)

Chúc bạn học tốt !

\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)

\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)

\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)

\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)

Chúc bạn học tốt!!!

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

\(\Rightarrow\left(x+1\right)^2\left(x+1\right)=8\)

\(\Rightarrow\left(x+1\right)^3=8\)

\(\Rightarrow\left(x+1\right)^3=2^3\)

\(\Rightarrow x+1=2\Rightarrow x=1\)