Giải

Một lít xăng đi được số ki-lô-mét là:

        10 + 10 = 20 (km)

75% của bình xăng là:

        5.75% = 3,75 (lít)

Còn 75% bình xăng thì đi được số ki-lô-mét là:

       3,75.20 = 75 (km)

Vậy 75% bình xăng đi được 75 km

 #hoktot<3# (dấu chấm là dấu nhân nhá ^^)

15.5 nha

COVID-19, a disaster! But while on vacation, I did very interesting things. I will tell you! When the epidemic was still raging, our students were able to study online via modern computers. As long as you stay at home and have a computer with an internet connection, you can meet anyone in the world. I get to eat cozy meals with my parents and brother every day, enjoy home-cooked meals, exercise more, read more books… Every day I bury my head in my studies, I forget. caring about people, I forgot my hobby, forgot about the lovely things around me. After a period of slowing down, I found my own value, the precious value of life. The first time I planted a seed, cared for it and watched it grow day by day, I was very happy. I told myself that everything will pass, life always has hope.

a) *) \(\frac{n-1}{3-2n}\)

Gọi d là ƯCLN (n-1;3-2n) (d\(\inℕ\))

\(\Rightarrow\hept{\begin{cases}n-1⋮d\\3-2n⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n-2⋮d\\3-2n⋮d\end{cases}\Leftrightarrow}\left(2n-2\right)+\left(3-2n\right)⋮d}\)

\(\Leftrightarrow1⋮d\left(d\inℕ\right)\Rightarrow d=1\)

=> ƯCLN (n-1;3-2n)=1

=> \(\frac{n-1}{3-2n}\)tối giản với n là số tự nhiên

*) \(\frac{3n+7}{5n+12}\)

Gọi d là ƯCLN (3n+7;5n+12) \(\left(d\inℕ\right)\)

\(\Rightarrow\hept{\begin{cases}3n+7⋮d\\5n+12⋮d\end{cases}\Rightarrow\hept{\begin{cases}15n+35⋮d\\15n+36⋮d\end{cases}\Leftrightarrow}\left(15n+36\right)-\left(15n+35\right)⋮d}\)

\(\Leftrightarrow1⋮d\left(d\inℕ\right)\)

\(\Rightarrow d=1\)

=> ƯCLN (3n+7;5n+12)=1

=> \(\frac{3n+7}{5n+12}\) tối giản với n là số tự nhiên

b) *) \(\frac{2n+5}{n-1}\left(n\ne1\right)\)

\(=\frac{2\left(n-1\right)+7}{n-1}=2+\frac{7}{n-1}\)

Để \(\frac{2n+5}{n-1}\) nhận giá trị nguyên => \(2+\frac{7}{n-1}\) nhận giá trị nguyên

2 nguyên => \(\frac{7}{n-1}\)nguyên

=> 7 chia hết cho n-1

n nguyên => n-1 nguyên => n-1\(\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Ta có bảng

vậy n={-6;0;2;8} thì \(\frac{2n+5}{n-1}\) nhận giá trị nguyên