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A= \(\frac{25^3.5^3}{6.5^{10}}\)= \(\frac{\left(5^2\right)^3.5^3}{6.5^{10}}\)= \(\frac{5^6.5^3}{6.5^{10}}\)= \(\frac{5^9}{6.5^{10}}\)= \(\frac{5}{6}\)
B = \(\frac{2^5.6^3}{8^2.9^2}\)= \(\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}\)=\(\frac{2^5.2^3.3^3}{2^6.3^4}\)= \(\frac{2^8.3^3}{2^6.3^4}\)= \(\frac{2^2}{3}\)= \(\frac{4}{3}\)
C = \(\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)= \(\frac{5^3.3^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}\)= \(\frac{5^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)= \(\frac{5^3.\left(1+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)= \(\frac{5^3.9}{6^3.35}\)
=\(\frac{5^3.3^2}{2^3.3^3.7.5}\)
= \(\frac{25}{168}\)
D = \(\frac{\left(7^4-7^3\right)^2}{49^3}\)= \(\frac{[7^3\left(7-1\right)]^2}{\left(7^2\right)^3}\)= \(\frac{7^6.6^2}{7^6}\)= \(36\)
a) (0.25)^3*32
= (0.5)^5*0.5*2^5
=1^5*0.5
=1*0.5
=0.5
b)(-0.125)^3*80^4
=(-0.125)^3*80^3*80
=(-0.125*80)^3*80
=(-10)^3
=-1000*80
=-80000
c) 8^2*4^5/2^20
=(23)2*(22)5*2^20
=2^6*2^10*2^20
=2^36
d)81^11*3^17/27^10*9^15
=((34)11*3^17)/(33)10*(32)15
=(3^44*3^17)/(3^30*3^30)
=3^61/3^60
=3
Bài b) dòng thứ 3 từ dưới đếm lên, phải là, (-10)^3*80 nha, gấp quá mình ghi nhầm ;)
c: \(=\dfrac{7}{23}\cdot\dfrac{-24-45}{18}=\dfrac{7}{23}\cdot\dfrac{-69}{18}=\dfrac{7}{18}\cdot\left(-3\right)=-\dfrac{7}{6}\)
d: \(=\dfrac{7}{5}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{9^5}=3^{40}-1\)
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
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) \(\sqrt{25}-\sqrt{\dfrac{49}{4}}+\sqrt{0,25}\)
\(=5-3,5+0,5\)
\(=2\)
b) \(\left(-\dfrac{25}{24}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\)
\(=\dfrac{-25}{24}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}\)
\(=\left(\dfrac{-25}{24}+\dfrac{7}{27}\right)+\left(-\dfrac{31}{42}+\dfrac{3}{42}\right)\)
\(=-\dfrac{169}{216}+-\dfrac{2}{3}=-1\dfrac{97}{216}\)
c) \(\dfrac{103}{10}-\left(9,5-0,25.18\right):0,5\)
\(=10,3-\left(9,5-0,25.18\right):0,5\)
\(=10,3-5:0,5\)
\(=0,3\)
d) \(\left(10^3+15.10^2+5^3\right):7\)
\(=\left(10^3+10^2\right)\left(15+5^3\right):7\)
\(=\left(1100.140\right):7\)
\(=22000\)