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Bài 1:
a. https://olm.vn/hoi-dap/detail/100987610050.html
b. Giống nhau hoàn toàn => P=Q
Chỉ biết thế thôi
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
Ta có : N = \(\frac{5.\left(2^3.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^6}{5.^{28}.3^{19}-7.2^{29}.3^{18}}\)
= \(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5-7.2\right)}\)
= \(\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
= \(\frac{3^{18}.2^{29}.\left(10-1\right)}{2^{28}.3^{18}.\left(-9\right)}=-2\)
Vậy N = -2
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5-7.2\right)}\)\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-14\right)}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}=\frac{2.9}{-9}=-2\)
Vậy A=-2
Ta có: \(A=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{28}\cdot3^{18}\cdot2}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2\cdot2^{28}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(5-7\cdot2\right)}\)
\(=\dfrac{2^{28}\cdot3^{18}\cdot\left(5\cdot2^2-2\right)}{2^{28}\cdot3^{18}\cdot\left(5-14\right)}\)
\(=\dfrac{20-2}{-9}=\dfrac{18}{-9}=-2\)
\(N=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^6}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(N=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(N=\frac{2^{28}.3^{18}\left(5.2^2-2.3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}\)
\(N=\frac{5.4-2.9}{5.3-7.2}=\frac{20-18}{15-14}=2\)
Vậy N = 2
=\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
=\(\frac{2^{29}\left(5.3^{18}-3^{18}\right)}{2^{28}\left(5.3^{18}-3^{18}\right).7}\)
=\(\frac{2}{7}\)
Bài 2: 1-5-9+13+17-21-25+29+...+2001-2005-2009+2013
=(1-5-9+13)+(17-21-25+29)+...+(2001-2005-2009+2013)
=0+0+...+0=0