`a, 3/4 - 5/4 :(x-1) =1/2`

`=> 5/4:(x-1)= 3/4 -1/2`

`=> 5/4:(x-1)= 3/4 - 2/4`

`=> 5/4:(x-1)= 1/4`

`=> x-1= 5/4 : 1/4`

`=> x-1=5`

`=>x=5+1`

`=>x=6`

__

`(1/2-x)^2 -2^2 =12`

`=> (1/2-x)^2 = 12+4`

`=> (1/2-x)^2= 16`

`=> (1/2-x)^2 =4^2`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)

__

`(1/2)^(2x-1) =1/16`

`=> (1/2)^(2x-1) = (1/2)^4`

`=> 2x-1=4`

`=> 2x=4+1`

`=>2x=5`

`=>x=5/2`

\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)

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

\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)

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

\(x-1=5\)

\(x=6\)

\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)

\(\left(\dfrac{1}{2}-x\right)^2-4=12\)

\(\left(\dfrac{1}{2}-x\right)^2=16\)

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

\(\dfrac{1}{2}-x=4hoặc\dfrac{1}{2}-x=-4\)

=>1/2 -x =4      1/2 -x= -4

=> x=1/2-4              x=1/2-(-4)

=>x=-7/2                 x=9/2

vậy x∈{-7/2 ; 9/2}

\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)

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

\(=>2x-1=4\)

\(=>2x=5\)

\(=>x=\dfrac{5}{2}\)

\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)

\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)

\(\Rightarrow x=\dfrac{2}{5}\)

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

\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)

\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)

\(\Rightarrow3x=\dfrac{7}{4}\)

\(\Rightarrow x=\dfrac{7}{4}:3\)

\(\Rightarrow x=\dfrac{7}{12}\)

\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)

\(\Rightarrow x=\dfrac{3}{4}\)

\(d,4^{x-3}+1=17\)

\(\Rightarrow4^{x-3}=17-1\)

\(\Rightarrow4^{x-3}=16\)

\(\Rightarrow4^{x-3}=4^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

#Toru

`3/2 x -1 =1/2x -3/5`

`=> 3/2x -1/2x = -3/5 +1`

`=> 2/2x= -3/5 + 5/5`

`=> x= 2/5`

__

`1/2x +1/2(x-2) = 3/4 -2x`

`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`

`=> 1/2x +1/2x +2x = 3/4 + 1`

`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`

`=> 6/2x = 7/4`

`=> x= 7/4 : 3`

`=>x=7/12`

__

`(x-1/2) -1/4=0`

`=> x-1/2=1/4`

`=> x=1/4 +1/2`

`=> x= 1/4 +2/4`

`=>x=3/4`

__

`4^(x-3) +1=17`

`=> 4^(x-3) =17-1`

`=> 4^(x-3)=16`

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

`=> x-3=2`

`=>x=2+3`

`=>x=5`

b) (5/2-3x)=25/9

            3x = 5/2-25/9

            3x =-5/18

              x =-5/18:3

              x=-5/54

\(e.\left(x-1\right)^5=-32\)

  \(\left(x-1\right)^5=\left(-2\right)^5\)

   \(x-1=-2\)

   \(x\)      \(=-2+1\)

   \(x\)        \(=-1\)

Vậy \(x=-1\)

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)

=>x=1 hoặc x=-1

\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)

\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)

\(=3^3-1+5^3-1\)

\(=27-1+125-1\)

\(=150\)

2 )

\(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

Vậy ...

b )

\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow4x=2\)

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

Vậy ...

c )

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

\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)

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

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

Vậy ...

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