Question

Giúp mik nhé !!

a,\(\left(x-3\right)\left(4-5x\right)=0\)

b,\(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)

c,\(5^x\cdot\left(5^3\right)^2=625\)

d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

Answer 1

d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\\ \Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\\ \Leftrightarrow x=\frac{6}{9}\)

Vậy...

Answer 2

a) \(\left(x-3\right).\left(4-5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{3;\frac{4}{5}\right\}.\)

b) \(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=0-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=-\frac{1}{3}.\)

Ta luôn có: \(\left|x\right|\ge0\) \(\forall x.\)

\(\Rightarrow\left|x+\frac{3}{4}\right|>-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|\ne-\frac{1}{3}.\)

Vậy \(x\in\varnothing.\)

c) \(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\)

\(\Rightarrow x+6=4\)

\(\Rightarrow x=4-6\)

\(\Rightarrow x=-2\)

Vậy \(x=-2.\)

Chúc bạn học tốt!

Answer 3

a) (x - 3) (4 - 5x) = 0

\(\Rightarrow\) \(\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=3\\5x=4\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=3\\x=0,8\end{matrix}\right.\)

Vậy x \(\in\) \(\left\{3;0,8\right\}\)

b) |x + \(\frac{3}{4}\) | + \(\frac{1}{3}\) = 0

|x + \(\frac{3}{4}\) | = \(\frac{1}{3}\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=\frac{-1}{3}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x+\frac{9}{12}=\frac{4}{12}\\x+\frac{9}{12}=\frac{-4}{12}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=\frac{-5}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy x \(\in\) \(\left\{\frac{-5}{12};\frac{-13}{12}\right\}\)

c) 5\(^x\) . (5\(^3\))\(^2\) = 625

5\(^x\) . 5\(^6\) = 5\(^4\)

x + 6 = 4

x = 4 - 6

x = -2

Vậy x = -2

d)(x - \(\frac{2}{9}\)) \(^3\) = (\(\frac{2}{3}\))\(^6\)

(x - \(\frac{2}{9}\)) \(^3\) = \([(\frac{2}{3})^2]^3\)

x - \(\frac{2}{9}\) = \(\frac{4}{9}\)

x = \(\frac{4}{9}\) + \(\frac{2}{9}\)

x = \(\frac{2}{3}\)

Vậy x = \(\frac{2}{3}\)