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a: Hai xe gặp nhau sau 360/(40+60)=3,6h=3h36'
Hai xe gặp nhau lúc:
7h+3h36'=10h36'
a: Hai xe gặp nhau sau
360/(40+60)=3,6h=3h36'
Hai xe gặp nhau lúc:
7h+3h36'=10h36'
\(2^{10}:64\cdot16\)
\(=2^{10}:2^6\cdot2^4\)
\(=2^{10-6+4}\)
\(=2^8\)
\(2^{10}.64.16\\ =2^{10}.2^6.2^4\\ =2^{10+6+4}=2^{20}\)
a: Ta có: \(\left(x-47\right)-115=0\)
\(\Leftrightarrow x-47=115\)
hay x=162
b: Ta có: \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow7x-11=10\)
\(\Leftrightarrow7x=21\)
hay x=3
c: Ta có: \(x^{10}=1^x\)
\(\Leftrightarrow x^{10}=1\)
hay \(x\in\left\{1;-1\right\}\)
d: Ta có: \(x^{10}=x\)
\(\Leftrightarrow x^{10}-x=0\)
\(\Leftrightarrow x\left(x^9-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e: Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
g: Ta có: \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot18=3^{14}\cdot2\)
Suy ra: \(3^x=3^{14}\)
hay x=14
\(\left(2x-133\right):3=33\\ 2x-133=99\\ 2x=232\\ x=116\)
Bài 1:
a; \(\dfrac{5}{4}\); \(\dfrac{-3}{25}\); \(\dfrac{7}{50}\)
4 = 22; 25 = 52; 50 = 2.52
BCNN(4; 25; 50) = 22.52 = 100
\(\dfrac{5}{4}\) = \(\dfrac{5.25}{4.25}\) = \(\dfrac{125}{100}\) ; \(\dfrac{-3}{25}\) = \(\dfrac{-3.4}{25.4}\) = \(\dfrac{-12}{100}\); \(\dfrac{7}{50}\) = \(\dfrac{7.2}{50.10}\) = \(\dfrac{14}{100}\)
b; \(\dfrac{-7}{20}\); \(\dfrac{9}{10}\); \(\dfrac{-13}{30}\)
20 = 22.5; 10 = 2.5; 30 = 2.3.5; BCNN(20;10;30) = 22.3.5 = 60
\(\dfrac{-7}{20}\) = \(\dfrac{-7.3}{20.3}\) = \(\dfrac{-21}{60}\); \(\dfrac{9}{10}\) = \(\dfrac{9.6}{10.6}\) = \(\dfrac{54}{60}\); \(\dfrac{-13}{30}\) = \(\dfrac{-13.2}{30.2}\) = \(\dfrac{-26}{60}\)
c; \(\dfrac{7}{16}\); \(\dfrac{13}{-18}\); \(\dfrac{-11}{-12}\)
16 = 24; 18 = 2.32; 12 = 22.3
BCNN(16; 18; 12) = 24.32 = 144
\(\dfrac{7}{16}\) = \(\dfrac{7.9}{16.9}\) = \(\dfrac{63}{144}\);
\(\dfrac{13}{-18}\) = \(\dfrac{-13}{18}\) = \(\dfrac{-13.8}{18.8}\) = \(\dfrac{-104}{144}\)
\(\dfrac{-11}{-12}\) = \(\dfrac{11}{12}\) = \(\dfrac{11.12}{12.12}\) = \(\dfrac{132}{144}\)