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\(A=\dfrac{2^4\cdot5^4\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)
\(=\dfrac{2^4\cdot3^2\cdot5^2\left(3^2\cdot5^2-1\right)}{2^8\cdot3^3\cdot5^2}=\dfrac{1}{16}\cdot\dfrac{1}{3}\cdot\dfrac{15^2-1}{1}\)
\(=\dfrac{224}{48}=\dfrac{14}{3}\)
A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
\(\dfrac{8^{10^{^{ }}}.15^{16}}{12^{15}.25^8}\)=\(\dfrac{2^{30}.5^{16}.3^{16}}{2^{30}.3^{15}.5^{16}}\)=3
\(\dfrac{4^{10}.9^6+3^{12}.8^5}{6^{13}.4-2^{16}.3^{12}}\)
\(=\dfrac{\left(2^2\right)^{10}.\left(3^2\right)^6+3^{12}.\left(2^3\right)^5}{\left(2.3\right)^{13}.2^2-2^{16}.3^{12}}\)
\(=\dfrac{2^{20}.3^{12}+3^{12}.2^{15}}{2^{13}.3^{13}.2^2-2^{16}.3^{12}}\)
\(=\dfrac{2^{20}.3^{12}+3^{12}.2^{15}}{2^{15}.3^{13}-2^{16}.3^{12}}\)
\(=\dfrac{2^{15}.3^{12}.\left(2^5+1\right)}{2^{15}.3^{13}.\left(3-2\right)}\)
\(=\dfrac{2^5+1}{3-2}\)
\(=\dfrac{32+1}{1}=33\)
1. a) (x-2)2 =1
=> x - 2 = \(\pm\sqrt{1}\)
=> x - 2 = 1 hoặc -1
=> x = 3 hoặc 1
b) 2x - 1= -8
=> 2x = -7
=>x = \(\dfrac{-7}{2}\)
c)thiếu đề
d) (x-1)x+2 = (x-1)x+4
(x-1)x+2 = (x-1)x+2+2
(x-1)x+2 = (x-1)x+2. (x-1)2
(x-1)x+2 - (x-1)x+2. (x-1)2 = 0
=> (x-1)x+2. [1 - (x-1)2] = 0
\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
b, \(A=\dfrac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^4}{2^{12}.3^6+\left(2^3\right)^4.3^5}=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}=\dfrac{2}{3.4}=\dfrac{1}{6}\)
a. Vì | x | = 1,5
\(\Rightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)
Với x = 1,5 ; y = -0,75 thì :
P = 1,5 - 4 . 1,5 . (-0,75) + ( -0,75 )
P = 1,5 - ( -4,5 ) -0,75
P = 6 - 0,75
P = 5,25
Với x = -1,5 ; y = -0,75 thì
P = -1,5 - 4 . ( -1,5 ) .( -0,75 ) + ( -0,75 )
P = -1,5 - 4,5 -0,75
P = -6 - 0,75
P = -6,75
\(\dfrac{12808125}{16}\)
Ý mình là viết cả cách làm ra