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
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Bài 2 :
\(a)\) Ta có :
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right)107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=1+\frac{1}{134.269+135}>1\)
Vậy \(A< B\)
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