Vậy câu hỏi là gì?
 

\(7^{1000}=\left(7^4\right)^{250}=\overline{......1}^{250}=\overline{.......1}\)

\(3^{1000}=\left(3^4\right)^{250}=\overline{......1}^{250}=\overline{......1}\)

\(\Rightarrow7^{1000}-3^{1000}=\overline{......1}-\overline{......1}=\overline{......0}⋮10\)

Ta có: \(A=100^2+200^2+300^2+...+1000^2\)

\(=100^2\cdot\left(1+2^2+3^2+...+10^2\right)\)

\(=100^2\cdot385=3850000\)

3800

a,3¹² và 5⁸

Ta có:3¹²=(3³)⁴=27⁴

5⁸=(5²)⁴=25⁴

Vì 27⁴>25⁴ nên 3¹²>5⁸

b,(0,6)⁹ và (0,9)⁶

Ta có:(0,9)⁶>(0,6)⁶ mà (0,6)⁶>(0,6)⁹

\(\Rightarrow\)(0,6)⁹<(0,9)⁶

c,5²⁰⁰⁰ và 10¹⁰⁰⁰

Ta có:5²⁰⁰⁰=(5²)¹⁰⁰⁰=25¹⁰⁰⁰>10¹⁰⁰⁰

\(\Rightarrow\)5²⁰⁰⁰>10¹⁰⁰⁰

d,?????

A = 2/3 + 2/3² + 2/3³ + ... + 2/3¹⁰⁰⁰

A/3 = 2/3² + 2/3³ + 2/3⁴ + ... + 2/3¹⁰⁰¹

-2A/3 = A/3 - A

= (2/3² + 2/3³ + 2/3⁴ + ... + 2/3¹⁰⁰¹) - (2/3 + 2/3² + 2/3³ + ... + 2/3¹⁰⁰⁰)

= 2/3¹⁰⁰¹ - 2/3

A = (2/3¹⁰⁰¹ - 2/3) : (-2/3)

= 1 - 1/3¹⁰⁰⁰

2 mũ 1000 > 12 mũ 1200

9 mũ 99 > 99 mũ 9

a, 12^1200 > 2^100 Vì cả cơ số lẫn số mũ đều lớn hơn

b, 9^99= (9^11)^9

Vì 9^11> 99 nêm 99^11^9> 99^9

Vậy 9^99> 99^9

\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)

\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)

\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)

\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)

\(g\left(x\right)=-2x^4+4x^3+x-5\)

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`

Bài 8:

a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)

b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)