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\(\frac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)
\(=\frac{2^{15}\cdot5^8+2^{14}\cdot5^9\cdot\left(-1\right)}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)
\(=\frac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}\)
\(=\frac{5\cdot\left(-3\right)}{2^2\cdot6}\)
\(=-\frac{5}{8}\)
\(=-\frac{5}{4}\)
\(8^x:2^x=16^{2017}\)
\(\Leftrightarrow\left(8:2\right)^x=16^{2017}\)
\(\Leftrightarrow4^x=16^{2017}\)
\(\Leftrightarrow4^x=\left(4^2\right)^{2017}\)
\(\Leftrightarrow4^x=4^{4034}\)
\(\Leftrightarrow x=4034\)
Vậy ....
Ta có : \(8^x:2^x=16^{2017}\Leftrightarrow2^{3x}:2^x=2^{4.2017}\Leftrightarrow2^{2x}=2^{8086}\Rightarrow2x=8086\Rightarrow x=4043\)
Vậy x =4043
a)2x+1+2x+2=48
=>2x.21+2x.22=48
=>2x.2+2x.4=48
=>2x.(2+4)=48
=>2x. 6 =48
=>2x = 48:6
=>2x = 8
=>2x = 23
=> x = 3
Vậy x = 3
Để ( x - 3 ) ( x3 - 8 ) ( x2 - 16 ) ( x2 - 9 ) = 0
thì x - 3 = 0
hoặc x3 - 8 = 0
hoặc x2 - 16 = 0
hoặc x2 - 9 = 0
→ x = 3
hoặc x3 = 8 = 23
→ x = 2
hoặc x2 = 16 = 42
→ x = 4
hoặc x2 = 9 = 32
→ x = 3
Vậy x = 3 ; x = 2 ; x = 4
Tổng các x là : 3 + 2 + 4 = 9
Bài 1:
a)
\(\dfrac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\\ =\dfrac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^{2\cdot2}\cdot5^{2\cdot2}+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4}{2^3\cdot5^2}+\dfrac{2^5\cdot5^3}{2^3\cdot5^2}\\ =2\cdot5^2+2^2\cdot5\\ =2\cdot25+4\cdot5\\ =50+20\\ =70\)
c)
\(\dfrac{\left(1-\dfrac{4}{9}-2\right)\cdot16}{\left(2-3\right)^{-2}}+12\\ =\dfrac{\left(\dfrac{9}{9}-\dfrac{4}{9}-\dfrac{18}{9}\right)\cdot16}{\left(-1\right)^{-2}}+12\\ =\dfrac{\dfrac{-13}{9}\cdot16}{\dfrac{1}{\left(-1\right)^2}}+12\\ =\dfrac{\dfrac{-208}{9}}{1}+12\\ =\dfrac{-208}{9}+12\\ =\dfrac{-208}{9}+\dfrac{108}{9}\\ =\dfrac{100}{9}\)
Bài 2:
a)
\(\left(x+2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
d) \(5^{\left(x-2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=2\)
\(2^{x^2+3}=2^{4.9}x2^{3.7}x2^{2.5}\Rightarrow2^{x^2+3}=2^{36}\cdot2^{21}\cdot2^{10}\Rightarrow2^{x^2+3}=2^{67}\)
=> x^2 + 3 = 67
=> x^2 = 65 => x =căn 65
Vì x nguyên => x = căn 65 không thỏa mãn
Vậy không có x
\(\dfrac{16^x}{8}=2^x\)
\(\Leftrightarrow16^x=2^x.8\)
\(\Leftrightarrow\left(2.8\right)^x=2^x.8\)
\(\Leftrightarrow2^x.8^x=2^x.8\)
\(\Leftrightarrow x=1\)