Giờ này còn mơ đẹp , ngủ ngon  gì nx

\(\left(x-5\right)^2+\left(3x-2\right)\left(3x+2\right)\)

\(=x^2-10x+25+9x^2-4\)

\(=10x^2-10x+21\)

\(3\cdot\left(x+\dfrac{1}{2}\right)=\dfrac{7}{4}-\dfrac{2}{3}\)

\(3\cdot\left(x+\dfrac{1}{2}\right)=\dfrac{21}{12}-\dfrac{8}{12}\)

\(3\cdot\left(x+\dfrac{1}{2}\right)=\dfrac{13}{12}\)

\(x+\dfrac{1}{2}=\dfrac{13}{12}\cdot\dfrac{1}{3}\)

\(x+\dfrac{1}{2}=\dfrac{13}{36}\)

\(x=\dfrac{13}{36}-\dfrac{1}{2}=\dfrac{13}{36}-\dfrac{18}{36}\)

\(x=-\dfrac{5}{36}\)

\(\left(2x-5\right)^2=81\)

\(4x^2-20x+25=81\)

\(4x^2-20x-56=0\)

\(4\cdot\left(x+2\right)\left(x-7\right)=0\)

\(\left[{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\Rightarrow x\in\left\{-2;7\right\}\)

\(\left(3x-4\right)^3=27\)

\(\left(3x-4\right)^3=3^3\)

`=>3x-4=3`

`3x=7`

`=>x=7/3`

\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4}-\sqrt{7}\right)^2\)

\(=8-2\sqrt{4+\sqrt{7}}\cdot\sqrt{4-\sqrt{7}}\)

\(=8-6\)

\(=2\)

\(\left[0,75-\dfrac{5}{6}\right]:\left|-\dfrac{4}{7}+\dfrac{1}{3}\right|\)

\(=\left[\dfrac{3}{4}-\dfrac{5}{6}\right]:\left|-\dfrac{4}{7}+\dfrac{1}{3}\right|\)

\(=\left[\dfrac{9}{12}-\dfrac{10}{12}\right]:\left|-\dfrac{12}{21}+\dfrac{7}{21}\right|\)

\(=-\dfrac{1}{12}:\dfrac{5}{21}\)

\(=-\dfrac{1}{12}\cdot\dfrac{21}{5}=-\dfrac{7}{20}\)

\(x^3=x\)

\(x^3-x=0\)

\(x\cdot\left(x+1\right)\cdot\left(x-1\right)=0\)

`=>`\(\left[{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

`=>`\(x\in\left\{0;-1;1\right\}\)

Bài 1 :

a) \(27^4:9^3=\left(3^{^3}\right)^4:\left(3^2\right)^3=3^{12}:3^6=3^6\)

b) \(\dfrac{6^2\cdot3^3}{12^2}=\dfrac{3^3}{4}=\dfrac{27}{4}\)

c) \(\dfrac{12^3\cdot18^2}{24^2}=\dfrac{12^3\cdot18^2}{12^2\cdot2^2}=12\cdot9^2=12\cdot81=972\)

Bài 2 :
a) \(4\cdot\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}=\left(-\dfrac{4}{2}\right)+\dfrac{1}{2}=-\dfrac{3}{2}\)

b) \(\left(\dfrac{3}{5}-\dfrac{3}{4}\right)\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}\right)^2\)

\(=\left(\dfrac{12}{20}-\dfrac{15}{20}\right)\cdot\left(\dfrac{5}{15}-\dfrac{3}{15}\right)^2\)

\(=\left(-\dfrac{3}{20}\right)\cdot\left(\dfrac{2}{15}\right)^2\)

\(=\left(-\dfrac{3}{20}\right)\cdot\dfrac{4}{225}=-\dfrac{1}{375}\)

\(99^{20}\) và \(99999^{10}\)

\(99^{20}=\left(99^2\right)^{10}=\left(9801\right)^{10}\)

mà \(\left(9801\right)^{10}< \left(99999\right)^{10}\)

`=>99^20<99999^10`

\(\)

\(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\dfrac{4}{15}\right|=-2,15+3,15\)

\(\left|x+\dfrac{4}{15}\right|=1,6=\dfrac{8}{5}\)

`=>`\(\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=-\dfrac{8}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=-\dfrac{8}{5}-\dfrac{4}{15}=-\dfrac{28}{15}\end{matrix}\right.\)

`=>`\(x\in\left\{\dfrac{4}{3};-\dfrac{28}{15}\right\}\)