Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}=\dfrac{1\cdot2}{3\cdot4}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=\dfrac{5\cdot\left(-6\right)}{1+8}=\dfrac{-30}{9}=\dfrac{-10}{3}\)
Câu 1:
2:
a: x/6-1/3=-3/2
=>x/6=-3/2+1/3=-9/6+2/6=-7/6
=>x=-7
b: =>|x+5|=7
=>x+5=7 hoặc x+5=-7
=>x=2 hoặc x=-12
1:
a: \(=1-\dfrac{5}{4}-\dfrac{3}{4}-2=-1-2=-3\)
b: \(=34\cdot\left(-420\right)-34\cdot580=34\cdot\left(-1000\right)=-34000\)
a: \(=\dfrac{-3}{7}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+2+\dfrac{3}{7}=2\)
b: \(=-\dfrac{5}{7}:\left(24-\dfrac{166}{7}\right)+\dfrac{37}{3}\)
\(=-\dfrac{5}{7}:\dfrac{2}{7}+\dfrac{37}{3}=\dfrac{-5}{2}+\dfrac{37}{3}=\dfrac{59}{6}\)
c: \(=4-\dfrac{32}{27}\cdot\dfrac{-27}{8}=4+4=8\)
d: \(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\dfrac{11+5}{20}\cdot\dfrac{5}{7}\)
\(=\dfrac{7}{5}-\dfrac{6}{20}\cdot\dfrac{5}{7}=\dfrac{29}{35}\)
\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{1}{6}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{21}{6}=\dfrac{7}{2}\)
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
Bài 2:
a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)
d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
2/3+(-2/3)=3/5+(3/-5)=0
i: 2/5-1/10=4/10-1/10=3/10
2/5+(-1/10)=4/10-1/10=3/10
5/6-2/3=5/6-4/6=1/6
5/6+(-2/3)=1/6
a: \(=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4\cdot\dfrac{12}{13}=\dfrac{-48}{13}\)
b: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{-35}{6}\cdot\dfrac{30}{31}-\dfrac{11}{31}=-6\)
5.
(x^2 -1)(x^2 +9) <0
(x+3)(x+1)(x-1)(x-3)<0
x \(\in\)(-3;-1)U(1;3)
1:
a: \(=\dfrac{6^5}{3^5}=2^5=32\)
b: \(=\dfrac{5^4\cdot2^4}{5^5\cdot\left(-2\right)^5}=\dfrac{1}{5}\cdot\dfrac{-1}{2}=\dfrac{-1}{10}\)
Bài 2:
a: \(\Leftrightarrow\left(\dfrac{1}{3}\right)^m=\left(\dfrac{1}{3}\right)^4\)
hay m=4
b: \(\Leftrightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^{10}\)
hay n=10