Ta có :

\(\left(2x+1\right)^3=27\)

\(\Rightarrow\left(2x+1\right)^3=3^3\)

\(\Rightarrow2x+1=3\)

\(\Rightarrow2x=3-1=2\)

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

Vậy \(x=1\)

Ta có : \(\left(2x+1\right)^3=27\)

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

2x + 1 = 3

2x = 2

x = 1

Vậy x = 1

\(\Rightarrow2\cdot x+1=3\\ \Rightarrow2\cdot x=2\\ \Rightarrow x=1\)

b: \(x\in\left\{4;-4\right\}\)

\(a,x^2=16\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ b,\left(2x-1\right)^3=-27\\ \Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\\ \Rightarrow2x-1=-3\\ \Rightarrow2x=-2\\ \Rightarrow x=-1\)

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

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

\(12x-33=3^2.3^3\)

\(\Rightarrow12x-33=3^5\)

\(\Rightarrow12x-33=243\)

\(\Rightarrow12x=243+33\)

\(\Rightarrow12x=276\)

    \(\Rightarrow x=276:12\)

    \(\Rightarrow x=23\)

tíc mình nha

2,  27.75+25.27-150

=27.(75+25) - 150

=27.100 -15.10

=27.10.10 - 15.10

=10.(27.10 -15)

=10.(270-15)

=10.255

=2550

1,   \(12x-33=\)\(^{3^2.3^3}\)

\(12x-33=9.27\)

\(12x-33=243\)

12x=243+33

12x=276

x=276 : 12

x=23

(-26)+4=-22

207-28:4-215=207-7-215=200-215=200+(-215)=-15

27.18+27.16-79.17=27.(18+16)-79.17=27.34-79.17=27.2.17-79.17=54.17-79.17=(54-79).17=[54+(-79)].17=-25.17=-425

15.2^3+8.11^2-8.36=15.8+8.121-8.36=8.(15+121-36)=8.100=800

\(32.27+32.53\)

\(=32\left(27+53\right)\)

\(=32.80\)

\(=2560\)

32 x 27 + 32 x 53

= 32 x ( 27 + 53)

= 32 x 80

=2560

32 * 27 + 32 * 53 = 32 * ( 27 + 53 )

= 32 *  80

= 2560

Ta có: \(\left(\left(\frac{2}{3}\right)^x\right)^3=\left(\frac{3}{2}\right)^3\)

=>   \(\left(\frac{2}{3}\right)^x=\frac{3}{2}\)

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

=>   x =  -1

\(\left(\frac{2}{3}^x\right)^3=\frac{27}{8}\)

\(\left(\frac{2}{3}^x\right)^3=\left(\frac{3}{2}\right)^3\)

=> \(\frac{2}{3}^x=\frac{3}{2}\)

\(\frac{2}{3}^x=\frac{2}{3}^{-1}\)

=> \(x=-1\)

ta có

\(\frac{3}{x-\frac{1}{2}}=\frac{x-\frac{1}{2}}{27}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=3.27\)

\(\Leftrightarrow x^2-x+\frac{1}{4}=81\)

\(\Leftrightarrow x^2-x-80,75=0\)

\(\Leftrightarrow4x^2-4x-323=0\)(nhân cả 2 vế với 4)

\(\Leftrightarrow4x^2-38x+34x-323=0\)

\(\Leftrightarrow2x\left(2x-19\right)+17\left(2x-19\right)=0\)

\(\Leftrightarrow\left(2x+17\right)\left(2x-19\right)=0\)

\(\Leftrightarrow\orbr{\orbr{\begin{cases}2x+17=0\\2x-19=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{17}{2}\\x=\frac{19}{2}\end{cases}}\)

vậy.....

15.(27+18+6)+15(23+12)=15((27+23)+(18+12)+6)                                                                                                                                                =15.(50+30+6)=15.86=1290