Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

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)

Thiên Hương đẹp quá đi mất?

 Cho hoi dap de hoi chi khong duoc noi lung tung day la pham loi trong hoi dap

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

1: A=2

=>\(\sqrt{x}+1=2\left(\sqrt{x}-2\right)\)

=>\(2\sqrt{x}-4=\sqrt{x}+1\)

=>\(\sqrt{x}=5\)

=>x=25

2: A<1

=>A-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>0<=x<4

3: A<1/3

=>A-1/3<0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{1}{3}< 0\)

=>\(\dfrac{3\sqrt{x}+3-\sqrt{x}+2}{3\left(\sqrt{x}-2\right)}< 0\)

=>\(\dfrac{2\sqrt{x}+5}{3\left(\sqrt{x}-2\right)}< 0\)

=>\(\sqrt{x}-2< 0\)

=>0<=x<4

4:
A=căn x

=>\(\sqrt{x}+1=x-2\sqrt{x}\)

=>\(x-3\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{13}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{3-\sqrt{13}}{2}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{11+3\sqrt{13}}{2}\)

\(a,x+\dfrac{1}{4}\times\dfrac{1}{5}=\dfrac{4}{5}\)

\(x+\dfrac{1}{20}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}-\dfrac{1}{20}\)

\(x=\dfrac{16}{20}-\dfrac{1}{20}\)

\(x=\dfrac{15}{20}=\dfrac{3}{4}\)

\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)

\(\dfrac{11}{3}:x=\dfrac{8}{3}-\dfrac{3}{2}\)

\(\dfrac{11}{3}:x=\dfrac{7}{6}\)

\(x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)

a: =>x+1/20=4/5

=>x=4/5-1/20=16/20-1/20=15/20=3/4

b: =>11/3:x=8/3-3/2=16/6-9/6=7/6

=>x=11/3:7/6=11/3*6/7=66/21=22/7

a)

`x+1/4xx1/5=4/5`

`x+1/20=4/5`

`x=4/5-1/20`

`x=16/20-1/20`

`x=15/20=3/4`

b)

`3 2/3:x=2 2/3-1 1/2`

`11/3:x=8/3-3/2`

`11/3:x=16/6-9/6`

`11/3:x=7/6`

`x=11/3:7/6`

`x=11/3xx6/7`

`x=22/7`

\(a,x+\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{4}{5}\)

\(\Rightarrow\)\(x+\dfrac{1}{20}=\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{4}{5}-\dfrac{1}{20}=\dfrac{16}{20}-\dfrac{1}{20}=\dfrac{3}{4}\)

\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)

\(\Rightarrow\dfrac{11}{3}:x=\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)