\(=\left(5x-2y\right)\left[\left(5x\right)^2+5x\cdot2y+\left(2y\right)^2\right]\)

=(5x)^3-(2y)^3

=125x^3-8y^3

Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

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

\(=x^2+3x-12\)

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

\(=-2x^2y^2-2xy+3xy+3\)

\(=-2x^2y^2+xy+3\)

d: \(x\left(xy-1\right)\left(xy+1\right)\)

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

\(=x^3y^2-x\)

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

Bài 3: 

Ta có: \(A=\left(x+2\right)\left(3x-1\right)-x\left(3x+3\right)-2x+7\)

\(=3x^2-x+6x-2-3x^2-9x-2x+7\)

=5

a: \(=\dfrac{2xy\left(2x^2y-4x+5\right)}{2xy}=2x^2y-4x+5\)

b: \(=\dfrac{x^2y\left(7x^2y-2y-5x^2y^3\right)}{3x^2y}=\dfrac{7}{3}x^2y-\dfrac{2}{3}y-\dfrac{5}{3}x^2y^3\)

25x+25= 125

25x      = 125-25

25x     = 100

    x     = 100: 25

    x      = 4

k mik nha

25x + 25 = 125

25x = 125 - 25

25x = 100

x = 100 : 25 

x = 4

25x+25=125

25x=125-25

25x=100

x=100:25

x=4

Vậy x=4

à nhầm x=1250

trẻ đú

x.(x+1)=2.(1+2+3+...+1250)

<=> x.(x+1)=2.(1250.1251)/2

<=>x.(x+1)=1250.1251

=>x=1251

\(S=1+2+2^2+...+2^{100}\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{101}\)

\(\Rightarrow S=2^{101}-1\)

\(\Rightarrow S=2^{101}-1< 2^{122}\)

S = 1 + 2 + 2^2 +......+ 2^100

2S = 2 x (1 + 2 + 2^2 +.......+ 2^100)

2S = 2 + 2^2 + 2^3 +....+ 2^100 + 2^101

2S - S = (2 + 2^2 + 2^3 +.....+2^100 + 2^101)-(1+2+2^2+.....+2^100)

S = 2^101 - 1

=> 2^101-1 < 2^122