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A=\(\frac{3^{10}+1}{3^9+1}\)>1
=> A=\(\frac{3^{10}+1}{3^9+1}\)> \(\frac{3^{10}+1+2}{3^9+1+2}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^{10}+3}{3^9+3}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3\left(3^9+1\right)}{3.\left(3^8+1\right)}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^9+1}{3^8+1}\)=B
vậy A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)
b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)
c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)
mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)
=> A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)A=54-53/53+54=1/107=2/214
B=135-133/134+135=2/169
tự so sánh tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2
a, TS= 54 . 107 -53=(53+1) .107-53=53.107+107-53=53.107+ 54
<=>
\(\frac{TS}{MS}\)=\(\frac{54.107+54}{54.107+54}\)=1
Bài 1 :
\(a)\) Gọi \(ƯCLN\left(n+1;2n+3\right)=d\)
\(\Rightarrow\)\(\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}}\)
\(\Rightarrow\)\(\left(2n+2\right)-\left(2n+3\right)⋮d\)
\(\Rightarrow\)\(2n+2-2n-3⋮d\)
\(\Rightarrow\)\(\left(-1\right)⋮d\)
\(\Rightarrow\)\(d\inƯ\left(-1\right)\)
Mà \(Ư\left(-1\right)=\left\{1;-1\right\}\)
\(\Rightarrow\)\(d\in\left\{1;-1\right\}\)
Do đó :
\(ƯCLN\left(n+1;2n+3\right)=\left\{1;-1\right\}\)
Vậy \(\frac{n+1}{2n+3}\) là phân số tối giản với mọi n
Chúc bạn học tốt ~
Ta có : \(A=\frac{3^{10}+1}{3^9+1}\) => \(A.\frac{1}{3}=\frac{3^{10}+1}{3^{10}+3}=\frac{\left(3^{10}+3\right)-2}{3^{10}+3}=1-\frac{2}{3^{10}+3}\)
\(B.\frac{1}{3}=\frac{3^9+1}{3^8+1}\Rightarrow B.\frac{1}{3}=\frac{3^9+1}{3^9+3}=\frac{\left(3^9+3\right)-2}{3^9+3}=1-\frac{2}{3^9+3}\)
Vì : \(\frac{2}{3^{10}+3}< \frac{2}{3^9+3}\) nên \(A>B\)
\(a,\frac{27}{82}< \frac{27}{83}=\frac{1}{3};\frac{26}{75}>\frac{25}{75}=\frac{1}{3}\)
nên\(\frac{27}{82}< \frac{26}{75}\)
\(b,\frac{49}{78}< \frac{52}{78}=\frac{2}{3};\frac{64}{95}>\frac{64}{96}=\frac{2}{3}\)
nên\(\frac{49}{78}< \frac{64}{95}\Rightarrow\frac{-49}{78}>\frac{64}{-95}\)
c, Rút gọn:\(\frac{2525}{2929}=\frac{25}{29};\frac{217}{245}=\frac{31}{35}\)
Ta có:\(1-\frac{25}{29}=\frac{4}{29};1-\frac{31}{35}=\frac{4}{35}\Rightarrow1-\frac{25}{29}>1-\frac{31}{35}\)
\(\Rightarrow\frac{25}{29}< \frac{31}{35}\)hay\(\frac{2525}{2929}< \frac{217}{245}\)
\(d,A=\frac{3^{10}+1}{3^9+1}=1+\frac{3}{3^9+1}\);\(B=\frac{3^9+1}{3^8+1}=1+\frac{3}{3^8+1}\)
Dễ dàng nhận thấy \(\frac{3}{3^9+1}< \frac{3}{3^8+1}\Rightarrow A< B\)
Xin lỗi bạn e, mk ko làm được. Chúc bạn học tốt