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\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
1) P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15
P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)
P=1/3-1/15= 4/15
2) a/ 0,2:1+3/5+80%
= 2/10:8/5+8/10
= 2/10.5/8+8/10
= 1/8 + 4/5 = 5/40 + 32/40 = 37/40
b/ 0,5:5/4-2+1/5
= 5/10:5/4-11/5
= 5/10.4/5-11/5
=2/5-11/5 = -9/5
Mình ko ghi lại đề nha
=>(-1/5)2x5-(26x38/37x26)
=1/5-3
=-14/5
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
\(N=-2\frac{2}{3}-0,2+1,75-\frac{5}{6}+0,7\)
\(N=\frac{-8}{3}-\frac{1}{5}+\frac{7}{4}-\frac{5}{6}+\frac{7}{10}\)
\(N=\frac{-160}{60}-\frac{12}{60}+\frac{105}{60}-\frac{50}{60}+\frac{42}{60}\)
\(N=\frac{\left[\left(-160\right)-12+105-50=42\right]}{60}\)
\(N=\frac{-75}{60}=\frac{-5}{4}\)
\(\Rightarrow N=\frac{-5}{4}\)
= -8/3 - 1/5 + 7/4 - 5/6 + 7/10 =
= -43/15 + 11/12 + 7/10 =
= -39/20 + 7/10 =
= -5/4
tk cho mình nhé bạn
\(a,51+6-1-0,2=55,8\)
\(b,2511\times-24,8-2511\times75,2\)
\(=-24,8\times75,2=-100\)
\(=-251100\)
a, 51+6-1-0,2
=57-1;2
=55.8
b, 2511. -24,8 -2511.75,2
=2511.(-24,8 -75,2)
=2511.-100
=-251100
a)\(0,2:1\frac{3}{5}+80\%=\frac{1}{8}+80\%=\frac{3205}{32}\)
b)\(0,5:\frac{5}{4}-2\frac{1}{5}=\frac{2}{5}-\frac{11}{5}=-\frac{9}{5}\)
\(0,2:1\frac{3}{5}+80\%=\left(\frac{2}{10}+\frac{8}{5}\right)+80\%=\left(\frac{20}{100}+\frac{160}{100}\right)+80\%=\frac{180}{100}+80\%=180\%+80\%=260\%\)
a) \(\frac{-2}{5}:\left(-0,2\right)=\frac{-2}{5}:\frac{-2}{10}=\frac{-2}{5}\cdot\left(-5\right)=\frac{\left(-2\right)\cdot\left(-5\right)}{5}=2\)
b) \(0,2:\frac{-2}{5}=\frac{2}{10}:\frac{-2}{5}=\frac{1}{5}\cdot\frac{-5}{2}=\frac{1\cdot\left(-5\right)}{5\cdot2}=\frac{-1}{2}\)
#Tokitou-Muichirou
a) \(-\frac{2}{5}:\left(-0,2\right)=-\frac{2}{5}\cdot\left(-5\right)=2\)
b) \(0,2:\left(-\frac{2}{5}\right)=\frac{2}{10}\cdot\left(-\frac{5}{2}\right)=-\frac{1}{2}\)