K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Gọi biểu thức trên là Acó:

A=1+1/2+1/2^2+1/2^3+...+1/2^99+1/2^100

2A=1/2+1/2^2+1/2^3+....+1/2^99+1/2^100+1/2^101

2A-A=(1/2+1/2^2+1/2^3+....+1/2^99+1/2^100+1/2^101)-(1+1/2+1/2^2+1/2^3+...+1/2^99+1/2^100)

A=1/2^101-1

A=-1

13 tháng 4 2022

rối quá :)

B = (-5)0 + 51 + (-5)2 + 53 + ... + (-5)2016 + 52017

B = 1 + 51 + 52 + 53 + ... + 52016 + 52017

5B = 5 + 52 + 53 + ... + 52016 + 52017

5B - B = (5 + 52 + 53 + ... + 52016 + 52017) - (1 + 51 + 52 + 53 + ... + 52016 + 52017)

   4B    =    52017                                          -       1

   B      =   \(\dfrac{5^{2017}-1}{4}\)

a: \(P=2x^2+3xy+y^2=\left(2x+y\right)\left(x+y\right)\)

\(=\left(2\cdot\dfrac{-1}{2}+\dfrac{2}{3}\right)\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{1}{6}=-\dfrac{1}{18}\)

d: \(Q=\dfrac{-1}{3}x^4y^2=\dfrac{-1}{3}\cdot16\cdot\dfrac{1}{16}=-\dfrac{1}{3}\)

18 tháng 4 2020

\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)

29 tháng 6 2017

a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)

\(=\left(\dfrac{13}{14}\right)^2\)

\(=\dfrac{169}{196}\)

b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=\left(\dfrac{-1}{12}\right)^2\)

\(=\dfrac{1}{144}\)

c,\(\dfrac{5^4.20^4}{25^5.4^5}\)

\(=\dfrac{100^4}{100^5}\)

\(=\dfrac{1}{100}\)

d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)

\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)

\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)

\(=4^4.\left(\dfrac{-10}{3}\right)\)

\(=256.\left(\dfrac{-10}{3}\right)\)

\(=\dfrac{-2560}{3}\)

23 tháng 9 2016

a.\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)

=\(\left(\frac{6}{14}+\frac{7}{14}\right)^2\)

=\(\left(\frac{13}{14}\right)^2\)

=\(\frac{13^2}{14^2}\)

=\(\frac{169}{196}\)

b.\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

=\(\left(\frac{9}{12}-\frac{10}{12}\right)^2\)

=\(\left(\frac{-1}{12}\right)^2\)

=\(\frac{-1^2}{12^2}\)

=\(\frac{1}{144}\).

c.Phần C bn viết lại đề bài đi,mk ko hiểu

d.\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)

=\(\frac{-10^5}{3^5}.\left(\frac{-6^4}{5^4}\right)\)

=\(\frac{-100000}{243}.\frac{1296}{625}\)

=\(\frac{-2560}{3}\)

                 Không biết đúng ko nữalolang

23 tháng 9 2016

c.\(\frac{5^4.20^4}{25^5.4^5}\)

=\(\frac{25^2.\left(5.4\right)^4}{25^5.4^5}\)

=\(\frac{1.\left(5\right)^4.\left(4\right)^4}{25^3.4^5}\)

=\(\frac{25^2.1}{25^3.4}\)

=\(\frac{1}{25.4}\)

=\(\frac{1}{100}\).

13 tháng 8 2019

3.

a) \(\left(x-1\right)^3=125\)

=> \(\left(x-1\right)^3=5^3\)

=> \(x-1=5\)

=> \(x=5+1\)

=> \(x=6\)

Vậy \(x=6.\)

b) \(2^{x+2}-2^x=96\)

=> \(2^x.\left(2^2-1\right)=96\)

=> \(2^x.3=96\)

=> \(2^x=96:3\)

=> \(2^x=32\)

=> \(2^x=2^5\)

=> \(x=5\)

Vậy \(x=5.\)

c) \(\left(2x+1\right)^3=343\)

=> \(\left(2x+1\right)^3=7^3\)

=> \(2x+1=7\)

=> \(2x=7-1\)

=> \(2x=6\)

=> \(x=6:2\)

=> \(x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

13 tháng 8 2019

Giúp mk với nha các bạn