Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=20^2-19^2+18^2-17^2+...+2^2-1^2\)
\(=\left(20+19\right)\left(20-19\right)+\left(18+17\right)\left(18-17\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=39+35+..+3\)
\(=210\)
\(\dfrac{8^2.4^5}{2^{20}}\) = \(\dfrac{\left(2^3\right)^2\left(2^2\right)^5}{2^{20}}\) = \(\dfrac{2^6.2^{10}}{2^{20}}\) = \(\dfrac{2^{16}}{2^{20}}\) = \(\dfrac{1}{2^4}\) = \(\dfrac{1}{16}\)
a) \(12^8.9^{12}=18^{16}\)
ta cóvế trái : \(12^8.9^{12}=\left(2^2.3\right)^8.\left(3^2\right)^{12}=2^{16}.3^8.3^{24}=2^{16}.3^{32}\)
vế phải :\(18^{16}=\left(2.3^2\right)^{16}=2^{16}.3^{32}\)
=> VT =VP
b) \(75^{20}=45^{10}.5^{30}\)
VT=\(75^{20}=\left(3.5^2\right)^{20}=3^{20}.5^{40}\)
VP = \(45^{10}.5^{30}=\left(2^2.5\right)^{10}.5^{30}=2^{20}.5^{30}\)
ta tháy VT=VP
=> ĐPCM
a)
\(\frac{16}{2^x}=2\)
\(\Rightarrow2^{x+1}=16\)
\(\Rightarrow2^{x+1}=2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b)
\(\frac{\left(-3\right)^x}{81}=-27\)
\(\Rightarrow\left(-3\right)^x=-\left(3^3.3^4\right)\)
\(\Rightarrow-3^x=-3^7\)
=> x=7
c)
\(8^n:2^n=4\)
\(\Rightarrow2^{3n}:2^n=4\)
\(\Rightarrow2^{3n-n}=4\)
\(\Rightarrow2^{2n}=2^2\)
=>2n=2
=>n=1
a)\(\frac{16}{2^n}=2\)
=>16:2n=2
=>2n=16:2
=>2n=8
b)ko nhớ cách làm
c)8n:2n=4
=>(23)n:2n=22
=>23n:2n=22
=>23n-n=22
=>22n=22
=>2n=2
=>n=1
dc rùi chứ
( 5/4x - 2/3 )^4 = 57/16 + 3/2
=> ( 5/4x - 2/3 )^4 = 81/16
=> ( 5/4x - 2/3 )^4 = 3^4/2^4
=> 5/4x - 2/3 = 3/2
=> 5/3x = 13/6
=> x = 13/10
Vậy x = 13/10
\(\left(\frac{5}{4}x-\frac{2}{3}\right)^4-\frac{3}{2}=\frac{57}{16}\)
\(\left(\frac{5}{4}x-\frac{2}{3}\right)^4-\frac{3}{2}=\frac{57}{16}+\frac{3}{2}\)
\(\left(\frac{5}{4}x-\frac{2}{3}\right)=\frac{81}{16}\)
Ta xét 2th:
Th1: \(\frac{5}{4}x-\frac{2}{3}=\left(\frac{81}{16}\right)^{\frac{1}{4}}\)
\(\Rightarrow x=\frac{26}{15}\)
Th2: \(\frac{5}{4}x-\frac{2}{3}=-\left(\frac{81}{16}\right)^{\frac{1}{4}}\)
\(\Rightarrow x=-\frac{2}{3}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{26}{15}\\x=-\frac{2}{3}\end{cases}}\)
S=1-1/4+1-1/9+...+1-1/x2
S=(1+1+1+...+1)-(1/4+1/9+...+1/x2)
Có (1/4+1/9+...+1/x2)<1/(1.2)+1/(2.3)+...+1/(x-1)x=1-1/x<1
=> (1/4+1/9+...+1/x2) ko là số nguyên
=>S ko là số nguyên
Ta có
\(N=2^2+4^2+...+16^2+18^2+20^2\)
\(=2^2\left(1+2^2+3^2+...+10^2\right)\)
Đặt \(A=1+2^2+3^2+...+10^2\)
= \(1+2\left(3-1\right)+3\left(4-1\right)+...+10\left(11-1\right)\)
= \(1+2.3+3.4+...+10.11-\left(2+3+...+10\right)\)
= \(B-54\)
LẠI CÓ:
\(B=1+2.3+3.4+...+10.11\)
=> \(3B=3+2.3.3+3.4.3+...+10.11.3\)
=> \(3B=3+2.3\left(4-1\right)+3.4.\left(5-2\right)+...+10.11.\left(12-9\right)\)
= \(3+2.3.4-1.2.3+3.4.5-2.3.4+...+10.11.12-9.10.11\)
= \(3-1.2.3+10.11.12\)
=> \(B=1-2+440=439\)
=> \(N=\left(439-54\right).4=385.4=1540\)