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18 tháng 10 2017

\(\dfrac{1}{3}-\left(1\dfrac{1}{2}-x\right)=0,3\\ \dfrac{1}{3}-\left(\dfrac{3}{2}-x\right)=\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{3}-\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{30}\\ x=\dfrac{3}{2}-\dfrac{1}{30}=\dfrac{44}{30}\)

18 tháng 10 2017

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!

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

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

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

18 tháng 10 2017

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!

18 tháng 10 2017

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

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

26 tháng 9 2017

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 !

28 tháng 9 2021

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
30 tháng 4 2022

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

AH
Akai Haruma
Giáo viên
30 tháng 4 2022

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$