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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 \(x_1=-3;x_2=2\)
a)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(3.2\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+3^8.2^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+3^8.2^{10}.5}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
b) đặt A=2100 - 299 + 298 - 297 +...+ 22 - 2
=>2A=2101-2100+299-298+...+23-22
=>2A+A=2101-2100+299-298+...+23-22+2100 - 299 + 298 - 297 +...+ 22 - 2
=>3A=2101-2
=>A=\(\frac{2^{101}-2}{3}\)