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

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

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

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

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

________

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

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

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

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

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

_____________

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

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

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

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

\(x=2\)

________

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

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           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=\(\dfrac{15}{4}\)}

a, (3 + x)^{2 = 4}

(3 + x)² = 2² = (-2)²

TH1: 3 + x = 2

        x = 2 - 3

       x = -1

TH2: 3 + x = -2

        x = -2 - 3

       x = -2 + (-3)

       x = -5

Vậy x ϵ {-1; -5}

b, (2x + 1) ³ = -8

(2x + 1)³ = (-2)³

2x + 1 = -2

2x = -2 - 1

2x = -2 + (-1)

2x = -3

x = \(\dfrac{-3}{2}\)

có ai ko

`#3107.101107`

\(\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{11}\times\dfrac{1}{7}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{7}\times\dfrac{1}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)

\(\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{11}\cdot\dfrac{1}{7}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{7}\cdot\dfrac{1}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}=\dfrac{-5}{11}\)

`(x-1)^3=1/8`

<=> `(x-1)^3=(1/2)^3`

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

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

<=> `x=3/2`

(x-1)³=1/8

(x-1)³=(1/2)³

x-1=1/2

x=3/2

`#3107.101107`

\(\dfrac{27^2\cdot2^3\cdot5^4}{15^2\cdot6^9}\)

\(=\dfrac{\left(3^3\right)^2\cdot2^3\cdot5^4}{3^2\cdot5^2\cdot2^9\cdot3^9}\)

\(=\dfrac{3^6\cdot2^3\cdot5^4}{3^{11}\cdot5^2\cdot2^9}\)

\(=\dfrac{5^2}{3^5\cdot2^6}\)

\(=\dfrac{25}{15552}\)

Đây nha:

\(\left(x-\dfrac{1}{8}\right)^3=-\dfrac{8}{125}\)

\(\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{-2}{5}\right)^3\)

\(\Rightarrow x-\dfrac{1}{8}=\dfrac{-2}{5}\)

\(x=\dfrac{-2}{5}+\dfrac{1}{8}=\dfrac{-16}{40}+\dfrac{5}{40}\)

\(x=\dfrac{-11}{40}\)

\(#WendyDang\)

\(\left(x-\dfrac{1}{3}\right)^3=\dfrac{-8}{125}\)

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

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

\(x=\dfrac{-2}{5}+\dfrac{1}{3}=\dfrac{-6}{15}+\dfrac{5}{15}\)

\(x=\dfrac{-1}{15}\)

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}\\ =\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}\\ =\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}\\ =\dfrac{2^2\cdot3}{2^6}\\ =\dfrac{4\cdot3}{64}\\ =\dfrac{12}{64}\\ =\dfrac{3}{16}\)