![](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)
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x}{6}=\frac{8}{3}\Rightarrow x=\frac{8\cdot6}{3}=16\)
b) \(\frac{-3}{x}=\frac{15}{7}\Rightarrow x=-\frac{3\cdot7}{15}=-\frac{7}{5}\)
c) \(\frac{0,1}{5}=\frac{x}{15}\Rightarrow x=\frac{15\cdot0,1}{5}=0,3\)
d) \(\frac{x-3}{3}=\frac{4}{5}\Leftrightarrow x-3=\frac{4\cdot3}{5}=\frac{12}{5}\)
\(\Rightarrow x=\frac{12}{5}+3=\frac{27}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,3(x-2)+4(x-3)-6(x+8)=15
3x-6+4x-12-6x+48-15=0
x-81=0
=>x=81
![](https://rs.olm.vn/images/avt/0.png?1311)
45^10*5^20/75^15
=5^10*9^10*5^20/(5^2)^15
=5^10*5^20*9^10/5^30
=9^10
(0.8)^5/(0.4)^6
=(0.4)^5*2^5/(0.4)^6
=2^5/(0.4)
=32/(0.4)
=80
2^15*9^4/6^6*8^3
=2^15*(3^2)^4/2^6*3^6*(2^3)^3
=2^15*3^8/2^6*3^6*2^9
=3^2
=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ thấy 80=79+1=x+1
Thay vào P(x) ta có:
\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+....+\left(x+1\right)x+15\)
\(P\left(x\right)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+....+x^2+x+15\)
\(P\left(x\right)=x+15=79+15=94\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left|x\right|=15-5=10\Leftrightarrow\orbr{\begin{cases}x=10\\x=-10\end{cases}}\)
\(b,\left|x-2\right|+3x=7\)
*) Với x<2\(pt\Leftrightarrow2-x+3x=7\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)(loại )
*) Với \(x\ge2\Leftrightarrow x-2+3x=7\)\(\Leftrightarrow4x=9\Leftrightarrow x=\frac{9}{4}\)(thỏa mãn x>=2)
\(c,\left|x-1\right|+2x=6\)
*) Với x<1\(pt\Leftrightarrow1-x+2x=6\Leftrightarrow x=5\)(loại)
*) Với\(x\ge1\Leftrightarrow x-1+2x=6\Leftrightarrow3x=5\)\(\Leftrightarrow x=\frac{5}{3}\)(thỏa mãn x>=1)
\(\text{a)}\left|x\right|+5=15\)
\(\Rightarrow\left|x\right|=10\)
\(\Rightarrow x=10\text{ hay }x=-10\)
\(\text{b)}\left|x-2\right|+3x=7\)
\(\Rightarrow\left|x-2\right|=7-3x\)
\(\text{TH1:}x-2\ge0\Rightarrow x\ge2\)
\(\Rightarrow x-2=7-3x\)
\(\Rightarrow x+3x=7+2\)
\(\Rightarrow4x=9\)
\(\Rightarrow x=\frac{9}{4}\left(\text{Nhận}\right)\)
\(\text{TH2:}x-2< 0\Rightarrow x< 2\)
\(\Rightarrow x-2=-\left(7-3x\right)\)
\(\Rightarrow x-2=-7+3x\)
\(\Rightarrow x-3x=-7+2\)
\(\Rightarrow-2x=-5\)
\(\Rightarrow x=\frac{5}{2}\left(\text{Loại}\right)\)
Vậy \(x=\frac{9}{4}\)
\(\text{c)}\left|x-1\right|+2x=6\)
\(\Rightarrow\left|x-1\right|=6-2x\)
\(\text{TH1:}x-1\ge0\Rightarrow x\ge1\)
\(\Rightarrow x-1=6-2x\)
\(\Rightarrow x+2x=6+1\)
\(\Rightarrow3x=7\)
\(\Rightarrow x=\frac{7}{3}\left(\text{Nhận}\right)\)
\(\text{TH2:}x-2< 0\Rightarrow x< 2\)
\(\Rightarrow x-1=-\left(6-2x\right)\)
\(\Rightarrow x-1=-6+2x\)
\(\Rightarrow x-2x=-6+1\)
\(\Rightarrow-x=-5\)
\(\Rightarrow x=5\left(\text{loại}\right)\)
Vậy \(x=\frac{7}{3}\)
cho m xin cái đề bài