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a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Ngô Hải Nam ơi bn trả lời giúp mik ik
bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên
`(2/3 x +1/2) (-2x+3)=0`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)\cdot\left(-2x+3\right)=0\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\dfrac{1}{4}\cdot\dfrac{1}{3}+\dfrac{1}{6}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{7}\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{7}\right)\\ =\dfrac{1}{4}\cdot\dfrac{9}{14}\\ =\dfrac{9}{56}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)
\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(A=\dfrac{1}{2^{2022}}-2\)
12A=12+(12)2+(12)3+(12)4+...+(12)202312A=12+(12)2+(12)3+(12)4+...+(12)2023
A−12A=(12)2023−1A−12A=(12)2023−1
12A=(12)2023−112A=(12)2023−1
A=122022−2
= 27/20 + -1/2 + 347/100 - 1/3 - 1/6
= 191/50