![](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)
a,x2.x3=25
=>x5=25
=>x=2
b,x+18=5.4^2
=>x=5.16-18
=>x=62
c,x.(x^2)^3=x^5
=>x.x5=x5
=>x=0,1
d,2x.7=224
2x=32
=>2x=25
=>x=5
e,(3x+5)2=289
=>(3x+5)2=172
=>3x+5=17
=>3x=12
=>x=4
g,32x+1.11=2673
=>32x=243
=>32x=35
=>x=\(\frac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{5}{2.1}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}.\left(\frac{1}{2}+\frac{1}{4}\right)\)
\(B=\frac{5}{2.1}+\frac{1}{11}.5\frac{1}{2}+\frac{1}{15}.\frac{3}{4}\)
tự giải tiếp nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
ta gọi phần trong ngoặc là A thì ta có
A nhân x = A
x= A-A
x=1
Đặt C = \(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)
=> 100C = \(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{10.110}\)
=> 100C = \(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\)
=> 100C = \(\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)
=> C = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}\)
Lại có B = \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
=> 10B = \(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\)
=> 10B = \(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\)
=> 10B = \(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\)
=> 10B = \(1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)
=> B = \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{10}\)
Khi đó \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
<=> C.x = B
<=> \(\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)}{100}x=\frac{1+\frac{1}{2}+...+\frac{1}{10}-\left(\frac{1}{101}+\frac{1}{102}+..+\frac{1}{110}\right)}{10}\)
=> \(x=10\)
Vậy x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a} x=1,0,,-1
b}x=2
\(x\times\left(x^2\right)^3=x^5\)
\(\Leftrightarrow x\times x^6=x^5\)
\(\Leftrightarrow x^7=x^5\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(3^{2x+1}\times11=2673\)
\(\Leftrightarrow3^{2x+1}=243\)
\(\Leftrightarrow3^{2x+1}=3^5\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)