a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

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

\(x=\dfrac{1}{5}\)

a: \(\Leftrightarrow\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{-3}{2}x+\dfrac{3}{4}\)

=>13/6x=3/4+5/6

=>13/6x=9/12+10/12=19/12

hay x=19/26

b: \(\left(5x-3\right)\left(2x+5\right)=0\)

=>5x-3=0 hoặc 2x+5=0

=>x=3/5 hoặc x=-5/2

c: \(\left(\dfrac{5}{6}:x-\dfrac{5}{4}\right)^4=\dfrac{81}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{33}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d: \(\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}-2=\dfrac{3}{2}\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}=\dfrac{3}{2}+2=\dfrac{7}{2}\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|=\dfrac{7}{2}:\dfrac{5}{4}=\dfrac{7}{2}\cdot\dfrac{4}{5}=\dfrac{28}{10}=\dfrac{14}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{5}=\dfrac{14}{5}\\\dfrac{2}{5}x-\dfrac{1}{5}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{2}{5}=\dfrac{15}{2}\\x=-\dfrac{13}{5}:\dfrac{2}{5}=\dfrac{-13}{2}\end{matrix}\right.\)

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a)

Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)

\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)

Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :

\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)

b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:

\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)

c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)

\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)

d) \((2x-1)^3=-27=(-3)^3\)

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

\(\Rightarrow 2x=-2\Rightarrow x=-1\)

e) \((x-2)^2=1=1^2=(-1)^2\)

\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)

f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)

\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)

g) \((x-1)^2=(x-1)^6\)

\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)

\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)

\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)

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

\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)

Vậy \(x=\left\{0;1;2\right\}\)

a)

\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

@Hùng Nguyễn

| 4x - 0,2 | = 0,2

=> 4x - 0,2 = \(\dfrac{-1}{5}\)

4x = \(\dfrac{-1}{5}\) + 0,2 = 0

x = \(\dfrac{0}{4}\)

=> Ko có giá trị x

=> 4x - 0,2 = \(\dfrac{1}{5}\)

4x = \(\dfrac{1}{5}\)+ 0,2 = \(\dfrac{2}{5}\)

x = \(\dfrac{2}{5}\): 4 = \(\dfrac{2}{5}\). 4 = \(\dfrac{8}{5}\)

Vậy x = \(\dfrac{8}{5}\)

a: Đặt A=0

=>-2/3x=5/9

hay x=-5/6

b: Đặt B(x)=0

=>(x-2/5)(x+2/5)=0

=>x=2/5 hoặc x=-2/5

c: Đặt C(X)=0

\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)

\(\Leftrightarrow x^3=-\dfrac{8}{27}\)

hay x=-2/3

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25