Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.....+\frac{1}{2^{99}}\)
\(\Rightarrow2^2A=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\)
\(\Rightarrow4A-A=2-\frac{1}{2^{99}}\)
\(\Rightarrow3A=2-\frac{1}{2^{99}}\)
\(\Rightarrow A=\frac{2-\frac{1}{2^{99}}}{3}\)
Giải:
\(A_5=\left(-2x^2+x-5\right)+2x\left(x-1\right)-\left(x-5\right)\)
\(\Leftrightarrow A_5=-2x^2+x-5+2x^2-2x-x+5\)
\(\Leftrightarrow A_5=\left(-2x^2+2x^2\right)+\left(x-x\right)+\left(-5+5\right)-2x\)
\(\Leftrightarrow A_5=-2x\)
Vậy ...
\(A_6=-2x^2\left(2-3x\right)-3x\left(2x^2+x-1\right)\)
\(\Leftrightarrow A_6=-4x^2+6x^3-6x^3-3x^2+3x\)
\(\Leftrightarrow A_6=\left(-4x^2-3x^2\right)+\left(6x^3-6x^3\right)+3x\)
\(\Leftrightarrow A_6=-7x^2+3x\)
Vậy ...
5A=5+52+53+...+550+551
5A-A=551-1
A=551-1:4
tick mk nha cái kia sai rôi
\(\frac{25^4\cdot7^2+5^8\cdot49}{5^8\cdot2^3-25^4}=\frac{5^8\cdot7^2+5^8\cdot7^2}{5^8\cdot2^3-5^8}=\frac{5^8\cdot7^2\cdot\left(1+1\right)}{5^8\cdot\left(2^3-1\right)}=\frac{2\cdot5^8\cdot7^2}{5^8\cdot7}=2\cdot7=14\)
Giải:
a) \(\dfrac{1}{3}\left(xy\right)^3.\left(-2\right)x.\dfrac{-3}{5}y^5z\)
\(=\dfrac{1}{3}.\left(-2\right).\dfrac{-3}{5}x^3y^3xy^5z\)
\(=\dfrac{2}{5}x^4y^8z\)
Vậy ...
b) \(-\dfrac{1}{3}x^2yz.\dfrac{1}{7}\left(xy\right)^4.\dfrac{7}{9}xyz^3\)
\(=-\dfrac{1}{3}.\dfrac{1}{7}.\dfrac{7}{9}x^2yz.x^4y^4.xyz^3\)
\(=-\dfrac{1}{27}x^7y^6z^4\)
Vậy ...
Ta có :A = 1 + 5 + 52 + 53 + .... + 549 + 550
=> 5A = 5 + 52 + 53 + .... + 550 + 551
=> 5A - A = 551 - 1
=> 4A = 551 - 1
=> A = \(\frac{5^{51}-1}{4}\)
A = 1 + 5 + 52 + 53 + ... + 549 + 550
5A = 5 + 52 + 53 + 54 + ... + 550 + 551
5A - A = (5 + 52 + 53 + 54 + ... + 550 + 551) - (1 + 5 + 52 + 53 + ... + 549 + 550)
4A = 551 - 1
\(A=\frac{5^{51}-1}{4}\)
A = 1 + 5 + 52 + 53 + ... + 549 + 550
5A = 5 + 52 + 53 + 54 + ... + 550 + 551
5A - A = (5 + 52 + 53 + 54 + ... + 550 + 551) - (1 + 5 + 52 + 53 + ... + 549 + 550)
4A = 551 - 1
$A=\frac{5^{51}-1}{4}$