Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(0,25.2\frac{1}{3}.30.0,5.\frac{8}{45}\)
\(=\frac{1}{4}.\frac{7}{3}.30.\frac{1}{2}.\frac{8}{45}\)
\(=\frac{14}{9}\)
2020/2021<1
2021/2022<1
2022/2023<1
2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023
=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4
\(20,23:0,5+20,23:0,25+20,23\times94\)
\(=20,23\times2+20,23\times4+20,23\times94\)
\(=20,23\times\left(2+4+94\right)=20,23\times100=2023\)
0,25 × X - 27/8 × X = 3/4
X × ( 0,25 - 27/8 ) = 0,75
X × ( -25/8 ) = 0,75
X = 0,75 ÷ ( -25/8 )
X = -6/25
\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)
\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(\left(4x-1\right)^3=27\)
⇔ \(\left(4x-1\right)^3=3^3\)
⇒ \(4x-1=3\)
⇔ \(x=1\)
\(3^{4x+1}=27^{x+3}\)
\(\Rightarrow3^{4x+1}=3^{3\left(x+3\right)}\)
\(\Rightarrow4x+1=3\left(3x+3\right)\)
\(\Rightarrow4x+1=3x+9\)
\(\Rightarrow4x-3x=9-1\) -> chuyển vế đổi dấu
\(\Rightarrow x=8\)
Vậy...
\(#NqHahh\)
\(0,5+0,25=\frac{3}{4}x+\frac{3}{4}\)
\(0,75=\frac{3}{4}x+\frac{3}{4}\)
\(\frac{3}{4}=\frac{3}{4}x+\frac{3}{4}\)
\(\frac{3}{4}x=\frac{3}{4}-\frac{3}{4}\)
\(\frac{3}{4}x=0\)
\(x=0:\frac{3}{4}=0\)
Vậy x = 0
ĐÚNG THÌ CHO MIK NHA