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a. \(1\frac{5}{7}\)-\(\frac{9}{7}\)*\(\frac{16}{9}\)
=\(\frac{12}{7}\)-\(\frac{16}{7}\)
=\(\frac{-4}{7}\)
b. \(\frac{-5}{8}\):\(\frac{1}{4}\)-\(\frac{6}{13}\)*4+\(\frac{3}{8}\)
=\(\frac{-5}{8}\cdot\)4-\(\frac{6}{13}\)*4+\(\frac{3}{8}\)
=4*(\(\frac{-5}{8}\)-\(\frac{6}{13}\))+\(\frac{3}{8}\)
=4*\(\frac{-113}{104}\)+\(\frac{3}{8}\)
=\(\frac{-113}{26}\)+\(\frac{3}{8}\)
=\(\frac{-413}{104}\)
c.( \(\frac{3}{8}\)+\(\frac{-1}{4}\)-\(\frac{5}{12}\)):\(\frac{1}{3}\)
=\(\frac{-7}{24}\)*3
=\(\frac{-7}{8}\)
Học tốt
a: 2/9-4/9=-2/9
b: 5/7-4/14=5/7-2/7=3/7
c: -2-5/8=-16/8-5/8=-21/8
d: =6/13+14/39=18/39+14/39=32/39
e: =4/5+4/18=72/90+20/90=92/90=46/45
f: =5/30-12/30=-7/30
g: =-63/63=-1
h: =4/13x26=4x2=8
a: 2/9-4/9=-2/9
b: 5/7 - 4/14 = 5/7 - 2/7 = 3/7
c: -2 - 5/8 = - 16/8 - 5/8 = - 21/8
d: = 6/13 + 14/39 = 18/39 + 14/39 = 32/39
e: = 4/5 + 4/18 = 72/90 + 20/90 = 92/90 = 46/45
f: = 5/30 - 12/30 = -7/30
g: = - 63/63 = -1
h: = 4/13 . 26 =4 . 2 = 8
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
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 >_< )
Lớp 6 chưa được học cái này mà
\(a^{n^{n^n}}\)
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