Tính \(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
5^102. 9^1009 / 3^2018 . 25^50
\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)
\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)
\(=\dfrac{5^2\cdot1}{1\cdot1}\)
\(=25\)
Thực hiện phép tính
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5.1}{1.1}=5\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)=\(\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}\) =5\(^2\) =25
1/ Tìm x , biết :
( 3x - 7 )2009 = ( 3x - 7 )2007
2/ Tính :
\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)
\(\Leftrightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)
\(\left(3x-7\right)^{2007}.\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\\left(3x-7\right)=\pm1\end{cases}}}\)
=> \(x=\frac{7}{3},x=2,x=\frac{8}{3}\)
Vậy ...
2/\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}=\frac{5^{100+2}.3^{2.1009}}{3^{2018}.5^{2.50}}=\frac{5^{100}.5^2.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
\(25.\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
Bài 1: Tính giá trị của biểu thức sau
A=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)
B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0
Mong mn giúp đỡ
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
Cho dãy số thực: \(a_1,a_2,a_3,............a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+...............+a^{2018}_{2018}=1009\). CM: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.........+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Cho dãy số thực: \(a_1,a_2,a_3,............a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+...............+a^{2018}_{2018}=1009\). CM: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.........+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Cho dãy số thực: \(a_1,a_2,a_3,.............,a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+.................+a_{2018}^{2018}=1009\). Chứng minh: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.............+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Bài 1: Thực hiện phép tính: (tính nhanh nếu có thể)
Câu 49. \(7\dfrac{3}{5}-\left(2\dfrac{5}{7}+5\dfrac{3}{5}\right)\)
Câu 50. \(\dfrac{-1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}\)
Giúp em 2 câu cuối với ạ <3
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
Câu 50 tus sửa đề
\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)
Bài 1 :
Câu 49 . = 38/5 - ( 19/7 + 28/5 )
= 38/5 - 19/7 - 28/5
= ( 38/5 - 28/5 ) - 19/7
= 2 - 19/7 = -5/7
Câu 50 . = ( -1/9 : -25/9 ) . 15/22
= 1/25 . 15/22 = 3/110
Đúng 100 % nha bạn 🤍