Ta có:

aa = 10 a + a =11a

bbb = 100b + 10b + b =111b

abab = 1000a+10a+100b+b=1010a + 101b

aabb=1000a+100a+10b+b=1100a+11b

Đáp số:

a)a^2

b)b^3

c)a^2,b^2

d)a^2,b^2

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

a: (7-x) mod 5

b: 20 div 9

a: (7-x) mod 5

b: 20 div 9

a: (7-x) mod 5

b: 20 div 9

lỗi

:v

 Bài 1 
\(\dfrac{6}{5}\)=\(1\dfrac{1}{5}\)
\(\dfrac{7}{3}\)=\(2\dfrac{1}{3}\)
\(-\dfrac{16}{11}\)=\(-1\dfrac{5}{11}\)
Bài 2 : 
\(5\dfrac{1}{7}\)=\(\dfrac{5.7+1}{7}\)=\(\dfrac{36}{7}\)( Dấu " . " là dấu nhân )
\(6\dfrac{3}{4}\)=\(\dfrac{6.4+3}{4}\)=\(\dfrac{27}{4}\)
\(-1\dfrac{12}{13}\)=\(-\dfrac{25}{13}\)
Bài 3 :

Chuyển phân số về hỗn số

Bài 4 :
Sorry Mình không biết làm:(
Bài 5 :
7%=\(\dfrac{7}{100}\)
45%=\(\dfrac{9}{20}\)
216%=\(\dfrac{54}{25}\)
Bài 6 :
\(\dfrac{3}{7}\)=\(\dfrac{7}{3}\)
\(6\dfrac{1}{3}\)=\(\dfrac{19}{3}\)=\(\dfrac{3}{19}\)
\(\dfrac{-1}{12}\)=\(\dfrac{-12}{1}\)
Bài 7 :
 

\(x^2y+3x-5\)

\(\rightarrow x\)*\(x\)*\(y+3\)*\(x-5\)

HĐT số 4: \(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)

__________

\(x^3+3x^2+3x+1\)

\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)

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

`x^3 +3x^2+3x+1`

`= x^3 + 3*x^2*1 +3*x*1^2 +1^3`

`=(x+1)^3`

x^3 + 3x^2 + 3x + 1

<=> x^3 + 3x^2.1 + 3x.1^2 + 1^3

<=> (x + 1) ^3