Câu 1: D

Câu 3: 53/144>9/170>9/230

Bài 3

\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)

\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)

\(=\dfrac{33}{23}\le x\le3\)

Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)

Có 1 giá trị thỏa mãn 

Chọn A

Bài 4

\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)

Chọn D

Bài 5

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(M=1-\dfrac{1}{100}\)

\(M=\dfrac{100}{100}-\dfrac{1}{100}\)

\(M=\dfrac{99}{100}\)

CHọn C

A
D
C

Nguyễn Việt Lâm; Nguyễn Lê Phước Thịnh giúp vs!

Gọi \(d=ƯC\left(n^2+4;n+5\right)\)

\(\Rightarrow n\left(n+5\right)-\left(n^2+4\right)⋮d\)

\(\Rightarrow5n-4⋮d\)

\(\Rightarrow5\left(n+5\right)-29⋮d\)

\(\Rightarrow29⋮d\)

\(\Rightarrow d=\left\{1;29\right\}\)

Phân số chưa tối giản \(\Leftrightarrow d\ne1\Rightarrow d=29\)

\(\Rightarrow n+5=29k\Rightarrow n=29k-5\)

\(1\le29k-5\le2020\Rightarrow\dfrac{6}{29}\le k\le\dfrac{2025}{29}\)

\(\Leftrightarrow1\le k\le69\Rightarrow\) có 69 số tự nhiên thỏa mãn

3 ?

\(\dfrac{13}{4}>x>\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{13}{4}>x>\dfrac{10}{4}\)

\(\Leftrightarrow x\in\left\{\dfrac{11}{4};\dfrac{12}{4}\right\}\)

\(\dfrac{-9}{7}+\dfrac{5}{-7}< x\le\dfrac{-5}{2}+\dfrac{18}{4}\)

\(\dfrac{-9}{7}+\dfrac{-5}{7}< x\le\dfrac{10}{4}+\dfrac{18}{4}\)

\(\dfrac{-14}{7}< x\le2\)

\(-2< x\le2\)

\(\Rightarrow\)\(x=\left\{-1;0;1;2\right\}\)

-2<-1,0,1,2<2

6(1-x)+4(2-x)<=3(1-3x)

=>6-6x+8-4x<=3-9x

=>-10x+14<=-9x+3

=>-x<=-11

=>x>=11

(1-2x)/4-2<-5x/8

=>2-4x-16<-5x

=>-4x-14<-5x

=>x<14

Số tự nhiên x thỏa mãn cả hai BPT khi và chỉ khi 11<=x<14

=>\(x\in\left\{11;12;13\right\}\)

mik cần gấp

D.2

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6