so sanh: A= \(\frac{3^{10}+1}{3^9+1}\) voi B=\(\frac{3^9+1}{3^8+1}\)
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gọi d là ƯC(7n + 4; 5n + 3)
=> 7n + 4 và 5n + 3 ⋮ d
=> 5(7n + 4) và 7(5n + 3) ⋮ d
=> 35n + 20 và 35n + 21 ⋮ d
=> (35n + 21) - (35n +20) ⋮ d
=> 1 ⋮ d
=> d = + 1
=> 7n+4/5n+3 là phân số tối giản
Gọi a C Ư(7n+4;5n+3)
=>7n+4 và 5n+3 đều chia hết cho a
=>5(7n+4) và 7(5n+3) chia hết cho a
=>35n+20 và 35n+21 chia hết cho a
=>(35n+21) - (35n+20) chia hết cho a
=>1chia hết cho a
=>d C { + 1 }
Vậy7n+45n+3 là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)
Tự so sánh
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)
Từ (1) và (2) suy ra 100A > 100B hay A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Do as directed
1. / / / // / // ( Rearrange these worlds in order)
->.....................My future house will be surrounded by tall trees and blue sea.............................
2. We tried our best, but we didn't win the game. ( Use Although)
-> Although..................We tried our best, we didn't win the game.....................
3. No mountain in Viet Nam is higher than Phanxipang. ( Use the Superlative of Adjective)
-> Phanxipang...........is the highest in VN...........................................
4. He goes to an English speaking club twice a week ( Make the question for underlined worlds)
->...........How often does he go to an English speaking club?......................................
5. We pollute the air. We have breathing problems. ( Use conditional sentence)
->.....................IF We didn't pollute the air,we wouldn't have breathing problems. ..............
6. Hung is putting his feet on the bench
-> You say:'' Don't..........put your feet on the beach,Hung.........................''
![](https://rs.olm.vn/images/avt/0.png?1311)
phân số chỉ số trang sách Mai đọc cả ngày thứ 2 & 3 là :
\(1-\frac{1}{4}=\frac{3}{4}\left(trang\right)\)
đổi :\(55\%=\frac{11}{20}\)
số trang sách đọc dc ngày thứ 2 chiếm :
\(\frac{3}{4}\cdot\frac{11}{20}=\frac{33}{80}\left(trang\right)\)
phân số chỉ ngày thứ 3 đọc dc là :
\(1-\frac{1}{4}-\frac{33}{80}=\frac{27}{80}\left(trang\right)\)
Cuốn sách có số trang là :
\(54:\frac{27}{80}=160\left(trang\right)\)
Đáp số : 160 trang
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi d là ƯC(7n + 4; 5n + 3)
=> 7n + 4 và 5n + 3 ⋮ d
=> 5(7n + 4) và 7(5n + 3) ⋮ d
=> 35n + 20 và 35n + 21 ⋮ d
=> (35n + 21) - (35n +20) ⋮ d
=> 1 ⋮ d
=> d = + 1
=> 7n+4/5n+3 là phân số tối giản
Đặt \(\left(7n+4;5n+3\right)=d\)
\(\Rightarrow\hept{\begin{cases}7n+4⋮d\\5n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}5.\left(7n+4\right)⋮d\\7.\left(5n+3\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}35n+20⋮d\\35n+21⋮d\end{cases}}\)
\(\Rightarrow\left(35n+21\right)-\left(35n+20\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow\frac{7n+4}{5n+3}\)là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
mình làm tắt nha
có số hs TB là
\(48.\frac{5}{12}=20\)(bạn)
có số hs giỏi là:
\(48-\orbr{ }\left(48-20\right).\frac{4}{7}+20\)=12 bạn
đáp số 12 bạn
\(A=\frac{3^{10}+1}{3^9+1}=\frac{3^{10}+3-2}{3^9+1}=\frac{3\left(3^9+1\right)-2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3^9+3-2}{3^8+1}=\frac{3\left(3^8+1\right)-2}{3^8+1}=3-\frac{2}{3^8+1}\)
Có \(3^9+1>3^8+1\)
\(\Rightarrow\frac{2}{3^9+1}< \frac{2}{3^8+1}\)
\(\Rightarrow3-\frac{2}{3^9+1}>3-\frac{2}{3^8+1}\)
\(\Rightarrow A>B\)