![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left|2x+1\right|< 2\)
\(\Rightarrow\left|2x+1\right|\in\left\{0;1\right\}\)
\(\Rightarrow2x+1\in\left\{-1;0;1\right\}\)
\(\Rightarrow2x\in\left\{-2;-1;0\right\}\)
\(\Rightarrow x\in\left\{-1;-\dfrac{1}{2};0\right\}\)
mà \(x\in Z\) nên \(x\in\left\{-1;0\right\}\)
b, \(11< x^2< 44\)
\(\Rightarrow x^2\in\left\{16;25;36\right\}\Rightarrow x\in\left\{-6;-5;-4;4;5;6\right\}\)
Vậy.......
Chúc bạn học tốt!!!
a) Ta có :
\(\left|2x+1\right|< 2\)
Mà \(\left|2x+1\right|\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|2x+1\right|=0\\\left|2x+1\right|=1\end{matrix}\right.\)
TH1 :
\(\left|2x+1\right|=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{2}\left(TM\right)\)
TH2 :
\(\left|2x+1\right|=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\2x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)\(\left(TM\right)\)
Vậy \(x\in\left\{\dfrac{-1}{2};0;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
a) \(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4^{2x-6}=1\)
\(\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
b) \(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
c) \(5< 5x< 125\)
\(\Rightarrow\frac{5}{5}< \frac{5x}{5}< \frac{125}{5}\)
\(\Rightarrow1< x< 25\)
\(\Rightarrow\left\{x\inℤ|1< x< 25\right\}\)
d) mk không hiểu
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3}{x}\ge1\)
\(\Leftrightarrow\frac{3}{x}\ge\frac{x}{x}\)
\(\Leftrightarrow3\ge x\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\) < vì x là các số nguyên dương>
VẬY \(x\in\left\{1;2;3\right\}\)
b) \(\frac{8}{x}< \frac{x}{3}< \frac{9}{x}\)
\(\Leftrightarrow\frac{24}{3x}< \frac{x^2}{3x}< \frac{27}{3x}\)
\(\Leftrightarrow24< x^2< 27\)
\(\Leftrightarrow\sqrt{24}< x< \sqrt{27}\)
\(\Leftrightarrow4,9< x< 5,2\) <Lưu ý căn 24 và căn 27 đã được làm tròn để viết cho gọn>
\(\Leftrightarrow x=5\) <vì x là số nguyên dương>
VẬY \(x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) 52-\(|\)x\(|\)=-80
\(|\)x\(|\)=52-(-80)
\(|\)x\(|\)=52+80
\(|\)x\(|\)=132
Vậy x=-132
a) \(|2x+1|< 2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1< 2\\-2x-1< 2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x>-\frac{3}{2}\end{cases}}\)
Vậy \(-\frac{3}{2}< x< \frac{1}{2}\)
11<x2<44<=>x2={16;25;36}<=>x={\(\pm\)4;\(\pm\)5;\(\pm\)6}
Vậy các giá trị nguyên của x là -6;-5;-4;4;5 và 6