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)
Vì (2x-1)^6=(2x-1)^8
(2x-1)^8-(2x-1)^6=0
(2x-1)^6[(2x-1)^2-1)]=0
th1 (2x-1)^6 suy ra 2x-1=0 suy ra x=1/2
th2 (2x-1)^2-1=0
(2x-1)^2=1
suy ra 2x-1 bằng 1;-1
th1 2x-1=1 suy ra x=1
2x-1=-1 suy ra x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Cách giải chung. Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\).
5. \(\frac{5a}{a+b}=\frac{5bk}{bk+b}=\frac{5k}{k+1}\)
\(\frac{5c}{c+d}=\frac{5dk}{dk+d}=\frac{5k}{k+1}\)
Suy ra đpcm.
6. \(\frac{a^2+3ab}{a^2-3b^2}=\frac{\left(bk\right)^2+3bk.b}{\left(bk\right)^2-3b^2}=\frac{k^2+3k}{k^2-3}\)
\(\frac{c^2+3cd}{c^2-3d^2}=\frac{\left(dk\right)^2+3dk.d}{\left(dk\right)^2-3d^2}=\frac{k^2+3k}{k^2-3}\)
Suy ra đpcm.
7, 8. Bạn làm tương tự.
![](https://rs.olm.vn/images/avt/0.png?1311)
3n + 3 + 3n + 1 + 2n + 3 + 2n + 2
= 3n.33 + 3n.3 + 2n.23 + 2n.22
= 3n.(27 + 3) + 2n.(8 + 4)
= 3n.30 + 2n.12
= 3n.5.6 + 2n.2.6
= 6.(3n.5 + 2n.2) \(⋮\) 6
Ta có: \(6-\left(x-\dfrac{1}{3}\right)^2=2^{2021}:\left(-2\right)^{2020}\)
\(\Leftrightarrow6-\left(x-\dfrac{1}{3}\right)^2=2^{2021}:2^{2020}\\ \Leftrightarrow6-\left(x-\dfrac{1}{3}\right)^2=2\\ \Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)