Bài 1: 

a) Ta có: \( \left|x+2\right|=\left|3-2x\right|\)

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

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

b) Ta có: \(\left|2x-4\right|=\left|3-x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=3-x\\2x-4=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+4\\2x-x=-3+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=7\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=1\end{matrix}\right.\)

mk ko bt

2^2.3^1-(1^2021+2021^2) : (-2)

= 4.3-(1+4084441 ) : (-2)

= 12-4084442:(-2)

= 12-(-2042221)

= 20421133

\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)

Vậy \(x=-\dfrac{2}{5}\)

x= \(\dfrac{7\pm\sqrt{37}}{3}\) nha

nhanh nhé mọi người

1 + 2 + 3 ... 2016

Day toan do co: ( 2016 - 1) : 1 + 1 = 2016 (so)

Vay tong cua bai toan la: ( 2016 + 1 ) . 2016 : 2 = 2033136

\(\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^2\right]^2\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^{2\cdot2}\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^4\)

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

\(=\left(\dfrac{1}{2}\right)^7\)

c: Gọi hai số cần tìm là x,y

Theo đề, ta có: \(9< 2x< y< 12\)

=>2x=10 và y=11

=>x=5 và y=11

d: Gọi hai số cần tìm là x,y

Theo đề, ta có: \(-15< 2x< y< -12\)

=>2x=-14 và y=-13

=>x=-7 và y=-13