3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

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

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

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

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

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

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

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

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

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

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

\(x=2\)

________

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

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

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

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

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

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

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

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

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

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

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

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

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

=>\(\dfrac{1}{4}:x=-\dfrac{5}{2}-\dfrac{3}{4}=-\dfrac{10}{4}-\dfrac{3}{4}=-\dfrac{13}{4}\)

=>\(x=\dfrac{-1}{4}:\dfrac{13}{4}=\dfrac{-1}{4}\cdot\dfrac{4}{13}=\dfrac{-1}{13}\)

b: \(\left(\dfrac{2}{3}\right)^{100}:x=\left(-\dfrac{2}{3}\right)^{98}\)

=>\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{2}{3}\right)^{98}\)

=>\(x=\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{2}{3}\right)^{98}=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)

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

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

=>\(\left[{}\begin{matrix}4x-\dfrac{1}{5}=2\\4x-\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{5}\\4x=-\dfrac{9}{5}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{11}{20}\\x=-\dfrac{9}{20}\end{matrix}\right.\)

\(\left(\dfrac{3}{4}x+5\right)-\left(\dfrac{2}{3}x-4\right)-\left(\dfrac{1}{6}x+1\right)=\left(\dfrac{1}{3}x+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Rightarrow\dfrac{3}{4}x+5-\dfrac{2}{3}x+4-\dfrac{1}{6}x-1=\dfrac{1}{3}x+4-\dfrac{1}{3}x+3\)

\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{2}{3}x-\dfrac{1}{6}x\right)+\left(5+4-1\right)=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(4+3\right)\)

\(\Rightarrow\left(\dfrac{9}{12}x-\dfrac{8}{12}x-\dfrac{2}{12}x\right)+8=7\)

\(\Rightarrow-\dfrac{1}{12}x+8=7\)

\(\Rightarrow-\dfrac{1}{12}x=-1\)

\(\Rightarrow x=12\)

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

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

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

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

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