![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{3^{10}.3^{10}.5^{10}.5^{10}.5^{10}}{3^{15}.5^{15}.5^{15}}\\ =\dfrac{3^{10+5}.3^{10-5}.5^{10+10+10}}{3^{15}.5^{15+15}}=\dfrac{3^{15}.3^5.5^{30}}{3^{15}.5^{30}}\\ =3^5=243\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=a^2+a+a^3-a+a=a^3+a^2+a=a\left(a^2+a+1\right)=VP\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, đề phải là 1/a.(a+1) = 1/a - 1/a+1 chứ bạn !
Có : 1/a.(a+1) = (a+1)-a/a.(a+1) = a+1/a.(a+1) - a/a.(a+1) = 1/a - 1/a+1
=> 1/a.(a+1) = 1/a - 1/a+1
b, Có : 2/a.(a+1).(a+2) = (a+2)-a/a.(a+1).(a+2) = a+2/a.(a+1).(a+2) - a/a.(a+1).(a+2) = 1/a.(a+1) - 1/(a+1).(a+2)
=> 2/a.(a+1).(a+2) = 1/a.(a+1) - 1/(a+1).(a+2)
Tk mk nha
a, \(VP=\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}==\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}=VT\)
b, \(VP=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}=VT\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của Nguyễn Quang Đức - Toán lớp 6 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn vào đây tìm đi Giáo án Toán 7 - Tuần 1 đến tuần 7 - Giáo Án, Bài Giảng
Đặt \(S=\frac{45^{10}\cdot5^{10}}{75^{15}}=\frac{\left(3^2\cdot5\right)^{10}\cdot5^{10}}{\left(3\cdot5^2\right)^{15}}\)
\(=\frac{3^{20}\cdot5^{10}\cdot5^{10}}{3^{15}\cdot5^{30}}=\frac{3^{15}\cdot3^5\cdot5^{10}\cdot5^{10}}{3^{15}\cdot5^{20}\cdot5^{10}}=\frac{3^5\cdot5^{10}}{5^{20}}=3^5\cdot5^{-10}=3^5\cdot\frac{1}{9765625}\)
Đến đây bạn tính nhé , đề cho sai phải không vậy?
Trả lời:
\(\frac{45^{10}.5^{10}}{75^{15}}=\frac{\left(3.15\right)^{10}.5^{10}}{75^{15}}\)
\(=\frac{3^{10}.15^{10}.5^{10}}{75^{15}}\)
\(=\frac{3^{10}.\left(15.5\right)^{10}}{75^{15}}\)
\(=\frac{3^{10}.75^{10}}{75^{15}}\)
\(=\frac{3^{10}}{75^5}\)
\(=\frac{3^{10}}{\left(3.25\right)^5}\)
\(=\frac{3^{10}}{3^5.25^5}\)
\(=\frac{3^5}{25^5}=\frac{243}{9765625}\)