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a) \(\frac{2^{47}\cdot5^{14}\cdot18^{13}\cdot45^{17}}{180^{30}}\)
=\(\frac{2^{47}\cdot5^{14}\cdot2^{13}\cdot3^{26}\cdot3^{34}\cdot5^{17}}{2^{60}\cdot3^{60}\cdot5^{30}}\)
=\(\frac{2^{60}\cdot3^{60}\cdot5^{31}}{2^{60}\cdot3^{60}\cdot5^{30}}\)
=\(\frac{5^{31}}{5^{30}}=5\)
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
a) \(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
= \(10.\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
= \(10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
= \(10.\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.\left(2.3-1\right)}\)
= \(10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
= \(10.\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)
= \(10.\dfrac{2^{12}.3^{10}.2.3}{2^{11}.3^{11}.5}\)
=\(10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)
= \(10.\dfrac{2^2.1}{1.1.5}\)
= \(10.\dfrac{4}{5}\)
= \(2.\dfrac{4}{1}\)
= 8
b) 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100
Đặt S = 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100
2S = 2 ( 1 + 2 + 22 + 23 + ... + 299 + 2100 )
= 2 + 22 + 23 + 24 + ... + 2100 + 2101
2S - S = ( 2 + 22 + 23 + 24 + ... + 2100 + 2101 ) - ( 1 + 2 + 22 + 23 + 24 + ... + 2100 )
= 2 + 22 + 23 + 24 + ... + 2100 + 2101 - 1 - 2 - 22 - 23 - 24 - ... - 2100
= 2101 - 1 = S
Vậy 1 + 2 + 22 + 23 + 24 + ... + 299 + 2100 = 2101 - 1