Question

Tìm x:

(4/9)^x+1=(8/27)^6

Answer 1

Trả lời :

\(\left(\frac{4}{9}\right)^{x+1}=\left(\frac{8}{27}\right)^6\)

\(\Rightarrow\left[\left(\frac{2}{3}\right)^2\right]^{x+1}=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\Rightarrow\left(\frac{2}{3}\right)^{2x+2}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x+2=18\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Answer 2

\(\left(\frac{4}{9}\right)^{x+1}=\left(\frac{8}{27}\right)^6=>\left[\left(\frac{2}{3}\right)^2\right]^{x+1}=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2.\left(x+1\right)}=\left(\frac{2}{3}\right)^{18}=>2.\left(x+1\right)=18\)

\(\left(x+1\right)=18:2=>\left(x+1\right)=9\)

\(x=9-1=>x=8\)

vậy \(x=8\)

Answer 3

\(\left(\frac{4}{9}\right)^{x+1}=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^{x+1}=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x+2}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x+2=18\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=16:2\)

\(\Rightarrow x=8\)

Học tốt

Answer 4

mình nhầm

\(\left(\frac{4}{9}\right)^{x+1}=\left(\frac{8}{27}\right)^6=>\left[\left(\frac{2}{3}\right)^2\right]^{x+1}=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\frac{2}{3}^{2.\left(x+1\right)}=\frac{2}{3}^{18}\)

\(\left(\frac{2}{3}\right)^{2.x+2}=\left(\frac{2}{3}\right)^{18}=>2.x+2=18\)

\(2.x=18-2=>2.x=16\)

\(x=16:2=>x=8\)

vậy \(x=8\)

Answer 5

mình làm hai bài nhưng bài nào cũng đúng nha còn chữ viết nhầm ở bài làm hai là do mình viết thêm thôi

Answer 6

           Bài làm :

Ta có :

\(\left(\frac{4}{9}\right)^{x+1}=\left(\frac{8}{27}\right)^6\)

\(\Leftrightarrow\left[\left(\frac{2}{3}\right)^2\right]^{x+1}=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x+2}=\left(\frac{2}{3}\right)^{18}\)

\(\Leftrightarrow2x+2=18\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Vậy x=8

Answer 7

cảm ơn mấy bạn nha!!!