a: \(27x^3-54x^2+36x-8=0\)

=>\(\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2-2^3=0\)

=>\(\left(3x-2\right)^3=0\)

=>3x-2=0

=>3x=2

=>\(x=\frac23\)

b: \(x^3+15x^2+75x+125=0\)

=>\(x^3+3\cdot x^2\cdot5+3\cdot x\cdot5^2+5^3=0\)

=>\(\left(x+5\right)^3=0\)

=>x+5=0

=>x=-5

c: \(x^3-18x^2+108x-216=0\)

=>\(x^3-3\cdot x^2\cdot6+3\cdot x\cdot6^2-6^3=0\)

=>\(\left(x-6\right)^3=0\)

=>x-6=0

=>x=6

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

=>\(x^3-3x^2+3x-1-x^3-3x^2=2\)

=>\(-6x^2+3x-3=0\)

=>\(2x^2-x+1=0\)

\(\Rightarrow x^2-\frac12x+\frac12=0\)

=>\(x^2-\frac12x+\frac{1}{16}+\frac{7}{16}=0\)

=>\(\left(x-\frac14\right)^2+\frac{7}{16}=0\) (vô lý)

=>x∈∅

e: \(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=3\)

=>\(x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3\)

=>15x-7=3

=>15x=10

=>\(x=\frac{10}{15}=\frac23\)

a) X  x  5  =  2435

               x = 2435 :  5

               x = 487

b) x  :  3  =  1075

            x = 1075 x  3

            x = 3225

a) X x 5 = 2435

    X       = 2435 : 5                                   

    X       = 487                                          

b) X : 3 = 1075

X     = 1075 x 3

X     = 3225

Bài 3:

a: \(\frac{x}{3}=\frac45\)

=>\(x=3\cdot\frac45=\frac{12}{5}\)

b: (x+2)(x-3)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=3\end{array}\right.\)

c: \(x^2-3x=0\)

=>x(x-3)=0

=>x=0 hoặc x=3

e: \(9^{x}=81\)

=>\(9^{x}=9^2\)

=>x=2

k: 2x=3y=5z

=>\(\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

=>\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

mà x+2y-z=29

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+2y-z}{15+2\cdot10-6}=\frac{29}{15+20-6}=\frac{29}{29}=1\)

=>x=15; y=10; z=6

(8796-1102).(8796-1102)+8796+1102

=7694.7694+8796+1102

=59197636+8796+1102 

=59207534

=>7694 x 7694 + 8796 + 1102 = 59207534

(k) đúng cho mình

(2-x)^3+(2+x)^3-12x(x+1)=0

=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)

=>\(12x^2+16-12x^2-12x=0\)

=>16-12x=0

=>4-3x=0

=>x=4/3

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)

\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)

\(x^2+6x+9-x^2+9=0\)

\(6x+18=0\)

\(6x=-18\)

\(x=-3\)

Vậy x=-3

\(b,5x^3+20x=0\)

\(5x\left(x^2+4\right)=0\)

\(Th1:5x=0=>x=0\)

\(Th2:x^2+4=0\)

\(x^2=-4\)(vô lý)

Vậy x=0

B3: 

a) x = 0

b) 9

B4 :

a) x = 1,2,3

b) x = 0,1,2

B5:

0,21; 0,215; 0,22

cho xin phép tính vs ạ