Question

[Mai Nguyễn]

Mọi người lm đc bao nhiêu thì lm ạ Nếu lm hết thì em rất cảm tạ🥺

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(\dfrac{1}{9}\cdot27^x=3^x\)

\(\Leftrightarrow\left(\dfrac{1}{9}\right)^x=\dfrac{1}{9}\)

hay x=1

b: Ta có: \(\left(3x+4\right)^2-8=41\)

\(\Leftrightarrow\left(3x+4\right)^2=49\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+4=7\\3x+4=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\3x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{3}\end{matrix}\right.\)

c: Ta có: \(\dfrac{\left(-3\right)^x}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4\)

Suy ra: x=7

Answer 2

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(\dfrac{1}{9}\cdot3^4\cdot3^x=3^x\)

\(\Leftrightarrow3^{x+2}-3^x=0\)

\(\Leftrightarrow x+2=x\left(loại\right)\)

b: Ta có: \(2\cdot\left(x+\dfrac{1}{4}\right)^3=-\dfrac{27}{4}\)

\(\Leftrightarrow x+\dfrac{1}{4}=-\dfrac{3}{2}\)

hay \(x=-\dfrac{7}{4}\)

c: Ta có: \(x:\left(-\dfrac{1}{3}\right)^2=-\dfrac{1}{3}\)

\(\Leftrightarrow x=-\dfrac{1}{3}\cdot\dfrac{1}{9}\)

hay \(x=-\dfrac{1}{27}\)

a: Ta có: \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)

\(\Leftrightarrow3^{2x}=\dfrac{1}{9}\)

\(\Leftrightarrow2x=-2\)

hay x=-1

b: Ta có: \(\dfrac{1}{4}+\left(2x-1\right)^3=\dfrac{1}{8}\)

\(\Leftrightarrow2x-1=-\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)