\(2012^0.\left(1,25-\dfrac{1}{4}\right)+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-\left|-5^2\right|\)
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TT
5 tháng 11 2017
a. \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
\(\Leftrightarrow\dfrac{31}{3}+\dfrac{4}{9}-\dfrac{26}{3}\)
\(=\dfrac{19}{9}\)
b. \(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}+\left(1,2.0,5\right):\dfrac{4}{5}\)
\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+0,6:\dfrac{4}{5}\)
\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
VN
Võ Ngọc Phương
VIP
11 tháng 7 2023
Tính hợp lí:
a) \(\left(-0,4\right)+\dfrac{3}{8}+\left(-0,6\right)\)
\(=\left[\left(-0,4\right)+\left(-0,6\right)\right]+\dfrac{3}{8}\)
\(=-1+\dfrac{3}{8}\)
\(=\dfrac{\left(-8\right)+3}{8}\)
\(=\dfrac{-5}{8}\)
b) \(\dfrac{4}{5}-1,8+0,375+\dfrac{5}{8}\)
\(=\dfrac{4}{5}-\dfrac{9}{5}+\dfrac{3}{8}+\dfrac{5}{8}\)
\(=-1+1\)
\(=0\\\)
c) \(\dfrac{7}{3}.\left(-2,5\right).\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{-5}{2}.\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{6}{7}.\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
d) \(\dfrac{7}{12}.\left(-2,34\right)-\dfrac{7}{12}.\left(-0,34\right)\)
\(=\dfrac{7}{12}.\left[\left(-2,34\right)+0,34\right]\)
\(=\dfrac{7}{12}.\left(-2\right)\)
\(=\dfrac{-7}{6}\)
e) \(\dfrac{-8}{3}.\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}\)
\(=\dfrac{8}{3}.\dfrac{-2}{11}-\dfrac{8}{3}.\dfrac{9}{11}\)
\(=\dfrac{8}{3}.\left(\dfrac{-2}{11}-\dfrac{9}{11}\right)\)
\(=\dfrac{8}{3}.-1\)
\(=\dfrac{-8}{3}\)
Chúc bạn học tốt
VN
Võ Ngọc Phương
VIP
7 tháng 9 2023
Bài 1:
a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)
b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=1\)
Bài 2:
a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)
b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)
c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)
d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)
Bài 3:
a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)
\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)
\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)
\(=9.\dfrac{2}{3}.-1\)
\(=6.-1=-6\)
b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
VN
Võ Ngọc Phương
VIP
7 tháng 9 2023
Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))
6 tháng 10 2021
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
6 tháng 10 2021
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
18 tháng 10 2021
\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)
\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)
\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
29 tháng 10 2021
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
NN
19 tháng 4 2017
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
PN
9 tháng 7 2018
\(=\dfrac{0,8:1}{0,6}+\dfrac{1:\dfrac{4}{7}}{3\dfrac{11}{36}.2\dfrac{2}{17}}+0,6:\dfrac{4}{5}\)
\(=\dfrac{0,4}{0,3}+\dfrac{\dfrac{7}{4}}{7}+0,75\)
\(=1\dfrac{0,4}{0,3}\)
\(2012^0.\left(1,25-\dfrac{1}{4}\right)+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-\left|-5^2\right|\)
\(=1,25-\dfrac{1}{4}+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-\left|-25\right|\)
\(=1,25-\dfrac{1}{4}+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)
\(=1,25-0,25+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)
\(=1+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)
\(=1+\left(\dfrac{3}{5}\right)^8.\left(\dfrac{3}{5}\right)^7-25\)
\(=1+\left(\dfrac{3}{5}\right)^{8+7}-25\)
\(=1+\left(\dfrac{3}{5}\right)^{15}-25\)
\(=1+\dfrac{3^{15}}{5^{15}}-25\)
\(=1,000470185-25\)
\(=-23,99952982\)
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