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

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

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

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

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

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

Bài 2:

a,\(|x-2|-3=-7\)

\(\Rightarrow|x-2|=-4\)

\(\Rightarrow x\in\varnothing\)

b,\(|x-5|+3=8\)

\(\Rightarrow|x-5|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)

c,\(|x+3|-2+2x=5\)

\(\Rightarrow|x+3|+2x=7\)

\(\Rightarrow\left[{}\begin{matrix}x+3=2x=7,x+3\ge0\\-\left(x+3\right)+2x=7,x+3< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=10,x< -3\end{matrix}\right.,x\ge-3\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x\in\varnothing\end{matrix}\right.\)

d,\(|x-3,1|+|y-1,2|=0\)

Thay y =0

\(\Rightarrow|x-3,1|+|0-1,2|=0\)

\(\Rightarrow|x-3,1|+|-1,2|=0\)

\(\Rightarrow|x-3,1|+1,2=0\)

\(\Rightarrow|x-3,1|=-1,2\)

\(\Rightarrow x\in\varnothing\)

Bài 3:

a,(-2,7)+9-3,5)-1,2(-2)

=>-2,7-3,5+2,4

=>-3,8

b, (-2,3)(-0,4)-(-3,1)(-2,1)-13,2

=>0,92+3,1(-2,1)-13,2

=>0,92-6,51-13,2

=-18,79

Bài 3 : Tính :

a) \(\left(-2,7\right)+\left(-3,5\right)-1,2.\left(-2\right)\)

\(=\left(-6,2\right)-\left(-2,4\right)\)

\(=\left(-6,2\right)+2,4=-3,8\)

b) \(\left(-2,3\right).\left(-0,4\right)-\left(-3,1\right).\left(-2,1\right)-13,2\)

\(=0,92-6,51-13,2\)

\(=-18,79\)

  \(1,2:\frac{3}{5}+\left(1\frac{1}{15}-\frac{8}{15}\right)x2\frac{2}{4}\)

\(=\frac{6}{5}x\frac{5}{3}+\left(\frac{16}{15}-\frac{8}{15}\right)x\frac{10}{4}\)

 \(=2+\frac{8}{15}x\frac{10}{4}=2+\frac{2}{3}\)

 \(=2\frac{2}{3}\)

\(1,2:\frac{3}{5}+\left(1\frac{1}{15}-\frac{8}{15}\right)\cdot2\frac{2}{4}\)

\(=\frac{6}{5}\cdot\frac{5}{3}+\left(\frac{16}{15}-\frac{8}{15}\right)\cdot\frac{10}{4}\)

\(=\frac{30}{15}+\frac{8}{15}\cdot\frac{10}{4}\)

\(=\frac{30}{15}+\frac{80}{60}=\frac{200}{60}\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

\(\left(x-1,2\right)^2=4\)

⇔\(x^2-2.x.1,2+1,2^2=4\)

⇔\(x^2-2,4x+1,44=4\)

⇔\(x^2-2,4x=4-1,44\)

⇔\(x\left(x-2,4\right)=2,56\)

⇔\(x=2,56\) hoặc \(x-2,4=2,56\)

⇔\(x=2,56\) hoặc \(x=4,96\)

a) \(\left(x-1,2\right)^2=4=2^2\)

\(\Leftrightarrow x-1,2=4\)

\(\Leftrightarrow x=5,2\)

b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=-6\)

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

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

⇒(\(x-1,2\))(\(x-1,2\))   =    8 \(\times\) 2

   (\(x-1,2\))²                =      16

    (\(x-1,2\))²                =       4²

    \(\left[{}\begin{matrix}x-1,2=4\\x-1,2=-4\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4+1,2\\x=-4+1,2\end{matrix}\right.\)

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

     Vậy \(x\) \(\in\) { -2,8; 5,2}

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

⇒ ( x - 1,2 )² = 8 . 2 = 16 = 4²

x - 1,2 = 4

x = 5,2

 \dfrac{x-1,2}{2}2x−1,2​                 =    \dfrac{8}{x-1,2}x−1,28​

⇒($x-1,2$)($x-1,2$)   =    8 $\times$ 2

   ($x-1,2$)2                =      16

    ($x-1,2$)2                =       42

    \left[{}\begin{matrix}x-1,2=4\\x-1,2=-4\end{matrix}\right.[x−1,2=4x−1,2=−4​

    \left[{}\begin{matrix}x=4+1,2\\x=-4+1,2\end{matrix}\right.[x=4+1,2x=−4+1,2​

    \left[{}\begin{matrix}x=5,2\\x=-2,8\end{matrix}\right.[x=5,2x=−2,8​
ii

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)

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

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{6}{5}-\dfrac{41}{5}\)

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{-36}{5}\) (vô lý vì \(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0\))

Vậy: Không tìm được giá trị x thoả mãn.