11×[(62+(-12)]+50+11

11×51+50+11

561+50+11

611+11

622

(5/13+8/13)+(-20/41+-21/41)+-5/7

1+-1+-5/7

-5/7

Ta có: A=\(\frac{20^8+1}{20^9+1}\)

=>20A=\(\frac{20^9+20}{20^9+1}\)=\(\frac{20^9+1+19}{20^9+1}=1+\frac{19}{20^9+1}\)

Lại có B=\(\frac{20^9+1}{20^{10}+1}\)

=>20B=\(\frac{20^{10}+20}{20^{10}+1}\)=\(\frac{20^{10}+1+19}{20^{10}+1}=\frac{20^{10}+1}{20^{10}+1}+\frac{19}{20^{10}+1}=1+\frac{19}{20^{10}+1}\)

Ta thấy \(20^9+1< 20^{10}+1\)

=>\(\frac{19}{20^9+1}>\frac{19}{20^{10}+1}\)

=>\(1+\frac{19}{20^9+1}>1+\frac{19}{20^{10}+1}\)

hay A>B
Vậy A>B

Xin lỗi vì sau 1 thời gian dài mới làm vì mik nghĩ bạn cx làm xong rồi nhưng coi như mik làm để tập quen vs nâng cao ik

Trả lời :

Ta có :

 \(\frac{x}{4}=\frac{1}{2}\Rightarrow2x=4\Rightarrow x=2\)

\(\frac{1}{2}=\frac{-y}{3}\Rightarrow2y=-3\Rightarrow y=\frac{-3}{2}\)

\(\frac{x}{4}=\frac{1}{2}=\frac{-y}{3}\)

* \(\frac{x}{4}=\frac{1}{2}\Rightarrow2x=4\Rightarrow x=2\)

* \(\frac{1}{2}=\frac{-y}{3}\Rightarrow3=-2y\Rightarrow y=-\frac{3}{2}\)

Bài làm:

a) Ta có: \(\hept{\begin{cases}4n-1⋮n-1\\n-1⋮n-1\end{cases}\Rightarrow\hept{\begin{cases}4n-1⋮n-1\\4n-4⋮n-1\end{cases}}}\)

\(\Rightarrow4n-1-\left(4n-4\right)⋮n-1\)

\(\Rightarrow3⋮n-1\)

\(\Rightarrow n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow n\in\left\{-2;0;2;4\right\}\)

Vậy \(n\in\left\{-2;0;2;4\right\}\)

b) Ta có: \(\hept{\begin{cases}n-1⋮n^2-2\\n^2-2⋮n^2-2\end{cases}\Rightarrow\hept{\begin{cases}n^2-n⋮n^2-2\\n^2-2⋮n^2-2\end{cases}}}\)

\(\Rightarrow n^2-2-\left(n^2-n\right)⋮n^2-2\)

\(\Rightarrow n-2⋮n^2-2\), mà ta có \(n-1⋮n^2-2\)

\(\Rightarrow n-1-\left(n-2\right)⋮n^2-2\)

\(\Rightarrow1⋮n^2-2\)

\(\Leftrightarrow n^2-2\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Leftrightarrow n^2\in\left\{1;3\right\}\)

Mà nếu n² = 3 thì n không là số nguyên

\(\Rightarrow n^2=1\Leftrightarrow\orbr{\begin{cases}n=1\\n=-1\end{cases}}\)

Vậy \(\orbr{\begin{cases}n=1\\n=-1\end{cases}}\)

Học tốt!!!!