BT1: Tìm x, biết:
1) \(\dfrac{1}{3}-\left(1\dfrac{1}{2}-x\right)=0,3\)
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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
\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)
\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)
\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)
\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)
Chúc bạn học tốt!!!
\(\dfrac{\left(x+1\right)^2}{2}=\dfrac{4}{x+1}\)
\(\Rightarrow\left(x+1\right)^2\left(x+1\right)=8\)
\(\Rightarrow\left(x+1\right)^3=8\)
\(\Rightarrow\left(x+1\right)^3=2^3\)
\(\Rightarrow x+1=2\Rightarrow x=1\)
\(\dfrac{1}{3}-\left(1\dfrac{1}{2}-x\right)=0,3\\ \dfrac{1}{3}-\left(\dfrac{3}{2}-x\right)=\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{3}-\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{30}\\ x=\dfrac{3}{2}-\dfrac{1}{30}=\dfrac{44}{30}\)