Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : 2x + 2x + 1 = 24
=> 2x(1 + 2) = 24
=> 2x.3 = 24
=> 2x = 8
=> 2x = 23
=> x = 3
Ta có : (x + 2)4 = (x + 2)6
=> (x + 2)4 - (x + 2)6 = 0
<=> (x + 2)4 (1 - (x + 2)2) = 0
<=> \(\orbr{\begin{cases}\left(x+2\right)^4=0\\\left(1-\left(x+2\right)^2\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\\left(x+2\right)^2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
a) x+2x+...+50x =2550
x. [ 1+2+3+....+50]=2550
ta co :
so so hang cua day 1;2;3;4;...;50:
[50-1]:1+1=50
tong cua day tren la :
[50+1].50:2=1275
=> x.1275=2550
x=2550:1275
vay x=2
a) \(\text{2(x-51)=2.2^2+20}\)
\(2\left(x-51\right)=2.4+20\)
\(2\left(x-51\right)=28\)
\(x-51=28\div2\)
\(x-51=14\)
\(x=14+51\)
\(\text{b)3.(x+1)-26=541}\)
\(3.\left(x+1\right)=541+26\)
\(3\cdot\left(x+1\right)=567\)
\(x+1=567\div3\)
\(x+1=189\)
\(x=189-1\)
\(x=188\)
\(x=65\)
\(\text{c)4(x-3)=7^2-1^10}\)
\(4\left(x-3\right)=49-1\)
\(4\left(x-3\right)=48\)
\(x-3=48\div4\)
\(x-3=12\)
\(x=12+3\)
\(x=15\)
\(\text{e)2x-138=2^3.3^2}\)
\(2x-138=8\cdot9\)
\(2x-138=72\)
\(2x=72+138\)
\(2x=210\)
\(x=210\div2\)
\(x=105\)
\(\text{f)(x-1)^4=16}\)
\(\left(x-1\right)^4=2^4\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
Bài 1:
\(=\left(15+47\right)\cdot42+42\cdot38=42\left(15+47+38\right)=42\cdot100=4200\)
Bài 2:
a: \(\Leftrightarrow3^x\left(1+3+3^2\right)=39\)
\(\Leftrightarrow3^x=3\)
hay x=1
b: \(\Leftrightarrow x^{2016}\left(1-x\right)=0\)
hay \(x\in\left\{0;1\right\}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\)
Đặt \(B=\frac{1}{3^2}+...+\frac{1}{2014^2}\)
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3}\)
.............
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)