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\(A=\frac{\left(-2\right)^0+1^{2017}+\left(-\frac{1}{3}\right)^8.3^8}{2^{15}}\)
\(=\frac{1+1+\frac{1}{3^8}.3^8}{2^{15}}\)
\(=\frac{1+1+1}{2^{15}}\)
\(=\frac{3}{2^{15}}\)
\(B=\frac{6^2}{2^{16}}\)
\(=\frac{2^2.3^2}{2^2.2^{14}}\)
\(=\frac{9}{2^{14}}\)
Dễ dàng thấy \(9>3\)
\(2^{14}< 2^{15}\)
Phép chia có cùng mẫu, tử lớn hơn thì đã lớn hơn, nay mẫu còn nhỏ hơn, chắc chắn rằng \(B>A\)
Vậy ...
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(a,\left[\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2\right]:\left[2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}\right)-\frac{27}{64}.4\right]:\left[2.\left(-1\right)+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}-\frac{27}{16}\right)\right]:\left[-2+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\frac{-2-27}{16}:\frac{-32+9-6}{16}\)
\(=-\frac{29}{16}:\frac{-29}{16}=1\)
\(b,\left[\left(\frac{4}{3}\right)^{-2}\left(\frac{3}{2}\right)^4\right]:\left(\frac{3}{2}\right)^6\)
\(=\left(\frac{9}{16}.\frac{81}{16}\right):\frac{729}{64}\)
\(=\frac{729}{64}:\frac{729}{64}=1\)
a, Tự chép đề bài ((:
\(=\frac{1}{9}\cdot1+\left(-\frac{1}{243}\right)\cdot\frac{9}{2}\)
\(=\frac{1}{9}-\frac{1}{54}\)
\(=\frac{5}{54}\)
b, 1. \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15}\)
\(=\left(\frac{2^5}{2^4\cdot2^4}\right)^5=\left(\frac{2^5}{2^8}\right)^5=\left(\frac{1}{2^3}\right)^5=\left(\frac{1}{8}\right)^5=\frac{1}{8^5}\)(Để vậy đi :v)
2. \(\left(\frac{2^6}{16^2}\right)^{10}\)
\(=\left(\frac{2^6}{2^8}\right)^{10}=\left(\frac{1}{2^2}\right)^{10}=\frac{1}{2^{20}}\)
c, \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)
\(=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^2}{1}=3^2=9\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
Ta có :
\(\frac{A}{B}=\frac{\left(-2\right)^0+1^{2017}+\left(\frac{-1}{3}\right)^8.3^8}{2^{15}}:\frac{6^2}{2^{16}}\)
=> \(\frac{A}{B}=\frac{1+1+\left(\frac{-1}{3}.3\right)^8}{2^{15}}.\frac{2^{16}}{6^2}\)
=> \(\frac{A}{B}=\frac{1+1+1^8}{1}.\frac{2}{6^2}\)
=> \(\frac{A}{B}=\frac{3}{1}.\frac{2}{2^2.3^2}\)
=> \(\frac{A}{B}=\frac{1}{2.3}=\frac{1}{6}\)
Ta có:
\(\frac{A}{B}\)=\(\frac{\left(-2\right)^0+1^{2017}+\left(\frac{-1}{3}\right)^8\cdot3^8}{2^{15}}\):\(\frac{6^2}{2^{16}}\)
=>\(\frac{A}{B}\)=\(\frac{1+1+\left(\frac{-1}{3}\cdot3\right)^8}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{1+1+1^8}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{3}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{2}{3.2^2}\)
=>\(\frac{A}{B}\)=\(\frac{1}{6}\)