Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a) Ta có số đối của 2,5 là -2,5

\(\Rightarrow x-3,5=-2,5\)

\(\Rightarrow x=-2,5+3,5\)

\(\Rightarrow x=1\)

b) Ta có số đối của -12 là 12

\(\Rightarrow3x-12=12\)

\(\Rightarrow3x=24\)

\(\Rightarrow x=\dfrac{24}{3}=8\)

c) Ta có số đối của \(-\dfrac{1}{8}\) là \(\dfrac{1}{8}\)

\(\Rightarrow2x+\dfrac{1}{4}=\dfrac{1}{8}\)

\(\Rightarrow2x=\dfrac{1}{8}-\dfrac{1}{4}\)

\(\Rightarrow2x=-\dfrac{1}{8}\)

\(\Rightarrow x=-\dfrac{1}{8}:2\)

\(\Rightarrow x=-\dfrac{1}{16}\)

d) Bạn viết lại đề

\(\dfrac{x}{2}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{20}\)

Áp dụng tc dstbn:

\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{20}=\dfrac{2x+3y-2z}{6\cdot2+3\cdot15-2\cdot20}=\dfrac{34}{17}=2\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=30\\z=40\end{matrix}\right.\)

Lời giải:

$\frac{x}{2}=\frac{y}{5}; \frac{y}{3}=\frac{z}{4}$

$\Rightarrow \frac{x}{6}=\frac{y}{15}=\frac{z}{20}$

Áp dụng TCDTSBN:

$\frac{x}{6}=\frac{y}{15}=\frac{z}{20}$

$=\frac{2x}{12}=\frac{3y}{45}=\frac{2z}{40}=\frac{2x+3y-2z}{12+45-40}=\frac{34}{17}=2$

$\Rightarrow x=2.6=12; y=2.15=30; z=2.20=40$

\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)

\(\Leftrightarrow2^x\cdot17=544\)

hay x=5

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)

       \(\frac{x+5}{3}=\frac{x-1}{4}\)

\(\Rightarrow\left(x+5\right).4=\left(x-1\right).3\)

\(\Rightarrow4x+20=3x-3\)

\(\Rightarrow4x-3x=-3-20\Rightarrow x=-23\)

\(\frac{x+5}{3}=\frac{x-1}{4}\)

\(\Rightarrow\left(x+5\right)\cdot4=\left(x-1\right)\cdot3\)

\(4x+20=3x-3\)

\(4x-3x=-3-20\)

\(x=-23\)

Vậy \(x=-23\)

\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+4}\)

\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+2}.\left(x-2\right)^2=0\)

\(\left(x-2\right)^{x+2}.\left[1-\left(x-2\right)^2\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^{x+2}=0\\1-\left(x-2\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=2\\x-2=1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=2\\x=3\end{cases}}\)

\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)

ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)

Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)

\(x-3,2+2x=x+\frac{16}{5}\)

\(x+2x=x+\frac{32}{5}\)

\(2x=\frac{32}{5}\)

\(\Leftrightarrow x=3,2\)(tm)

\(x-3,2+2x-\frac{1}{5}=3-x\)

\(x-3,2+2x=3-x+\frac{1}{5}\)

\(x-3,2+2x=\frac{16}{5}-x\)

\(x+2x=\frac{16}{5}-x+3,2\)

\(x+2x=\frac{32}{5}-x\)

\(2x=\frac{32}{5}-x-x\)

\(2x=\frac{32}{5}-2x\)

\(4x=\frac{32}{5}\)

\(x=1,6\)(tm)

Vậy \(x=1,6\)hoặc \(x=3,2\)