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MV
10 tháng 6 2017
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)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
Q
10 tháng 6 2017
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 \(x_1=-3;x_2=2\)
MC
3 tháng 8 2018
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
D
18 tháng 5 2016
1.a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\frac{3^{15}}{5^{15}}.\frac{5^{10}}{3^{10}}=\frac{3^5}{5^5}=\left(\frac{3}{5}\right)^5\)
b)\(\left(\frac{2}{3}\right)^{10}:\left(\frac{4}{9}\right)^4=\frac{2^{10}}{3^{10}}.\frac{3^8}{2^8}=\frac{2^2}{3^2}=\left(\frac{2}{3}\right)^2\)
2.
a)\(2^x=4\Rightarrow2^x=2^2\Rightarrow x=2\)
b)\(x^3=-27\Rightarrow x^3=-3^3\Rightarrow x=-3\)
c)\(x^2=16\Rightarrow x=\pm4\)
d)\(\left(x+1\right)^2=9\Rightarrow\hept{\begin{cases}x+1=3\Rightarrow x=2\\x+1=-3\Rightarrow x=-4\end{cases}}\)
CP
1
31 tháng 1 2017
a/\(^{2867971991^{-10}}\)
b/\(^{1125899907^{-55}}\)
c/\(^{1693508781^{-05}}\)
d/\(^{2867971991^{-10}}\)
\(a,\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\frac{1}{16^3}.8^2=\frac{\left(2^3\right)^2}{\left(2^4\right)^3}=\frac{2^{3.2}}{2^{4.3}}=\frac{2^6}{2^{12}}=\frac{1}{2^6}\)
\(b,\left(\frac{1}{9}\right)^{25}:\left(\frac{1}{3}\right)^{30}=\frac{1}{9^{25}}.3^{30}=\frac{3^{30}}{\left(3^2\right)^{25}}=\frac{3^{30}}{3^{50}}=\frac{1}{3^{20}}\)