1- height
2- cheerfulness
3- memorizes
4- forceful/ indecisive
5- unendurable
6- loss
7- laughter
8- socially
9- enjoyable
10-youthful

1. Did you ever be

2.Did you go

3.went

4.got up

5.never get up

6.went

7.Have you ever seen

8.was go

\(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\\ =\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{3}{4}\)

a)\(A=x^5-100x^4+100x^3-100x^2+100x-9\)

\(A=x^5-(99+1)x^4 +(99+1)x^3-(99+1)x^2+(99+1)x-9\)

Tại x=99 , ta có :

\(A=x^5 - (x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-9\)

\(A=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)

\(A=x-9\)

Thay x = 99 vào biểu thức A ta có :

\(A=99-9=90\)

Ta có : \(7^{1992}=\left(7^4\right)^{498}\)
Mà .
Vậy có 2 chữ số tận cùng là 01.

a)Ta có : \(10^{20}=\left(10^2\right)^{10}=100^{10}\)

Vì \(100^{10}>90^{10}\Rightarrow10^{20}>90^{10}\)

b)Ta có: \(\left(-5\right)^{30}=\left[\left(-5\right)^2\right]^{15}=25^{15}\)

\(\left(-3\right)^{30}=\left[\left(-3\right)^2\right]^{15}=9^{15}\)

Vì \(25^{15}>9^{15}\Rightarrow\left(-5\right)^{30}>\left(-3\right)^{30}\)

c)Ta có :\(32^9=\left(2^5\right)^9=2^{45}>2^{42}=\left(2^6\right)^7=64^7\\ 6^{13}< 6^{14}=\left(6^2\right)^7=36^7\)

Vì \(2^{45}>64^7>36^7>6^{13}\Rightarrow2^{45}>6^{13}\Rightarrow32^9>6^{13}\)

Vậy \(\left(-32\right)^9>\left(-6\right)^{13}\)

Sorry cop nhầm Câu hỏi của nguyen thi thien lam - Toán lớp 7 | Học trực tuyến