\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)

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

\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)

\(\Rightarrow5x-6=\dfrac{1}{4}\)

\(\Rightarrow5x=\dfrac{1}{4}+6\)

\(\Rightarrow5x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:5\)

\(\Rightarrow x=\dfrac{5}{4}\)

b) \(4^{x+2}+4^x=272\)

\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)

\(\Rightarrow4^x\cdot\left(16+1\right)=272\)

\(\Rightarrow4^x\cdot17=272\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)

\(\left|\dfrac{1}{2}x\right|=3-2x\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\\\dfrac{1}{2}x=-3+2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\)

`2:(1/2-2/3)^2+0,125^3xx8^3-(-12)^4:36^2`

`=2:(-1/6)^2+(0,125*8)^3-12^4:6^4`

`=2:1/36+1-(12:6)^4`

`=2xx36+1-2^4`

`=72+1-16`

`=73-16`

`=57`

2:(\(\dfrac{1}{2}\)-\(\dfrac{2}{3}\))\(^2\)+0,125\(^3\)x8\(^3\)-(-12)\(^4\):36\(^2\)

=2:(\(\dfrac{-1}{6}\))\(^2\)+\(\dfrac{1}{512}\)x512+12\(^4\):1296

=2:\(\dfrac{1}{36}\)+\(\dfrac{1}{512}\)x512+20736:1296

=72+1+16

=73+16

=89

\(2x-3\text{​​}\text{ ⋮ }x-1\)

\(\Rightarrow2\left(x-1\right)-1\text{ ⋮ }x-1\)

\(\text{mà 2(x-1) ⋮ x-1 }\)\(\Rightarrow1\text{ ⋮ }x-1\)

\(\Rightarrow\)\(x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Vậy \(x\in\left\{0;2\right\}\)

Đoạn \ldots là sao nhỉ? 

giải nhanh hộ mik nha

Không bình luận không liên quan đến nội dung của câu hỏi, làm phiền bạn tus.

Lời giải:

$|1-\sqrt{23}|+23-\sqrt{23}-|-2023|^0=\sqrt{23}-1+23-\sqrt{23}-1$

$=23-2=21$