Rút gọn phân số
a)\(\frac{-360}{450}\)
b)\(\frac{-260}{-1500}\)
c)\(\frac{2.3.5.13}{26.35}\)d)\(\frac{18.6-18}{\left(-36\right)\left(-5\right)}\)E)\(\frac{6.4+6.7}{6.5+12}\)F)\(\frac{25.9-25.17}{-8.80-8.10}\)g)\(\frac{^{3^2.5-3^6}}{3^4.13-3^4}\)
K
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3 tháng 3 2018
a)2.3.5.13/26.35
=2.3.5.13/2.13.5.7
=3/7
b)18.6-18/-36.(-35)
=18.6-18/-(36.25)=18(6-1)/-(36.35)
=18.5/-(36.35)=2.3.3.5/-(2.2.3.3.5.7)
=-1/14
c)6.4+6.7/6.5+12=6.4+6.7/6.5+6.2
=6(4+7)/6(5+2)
=11/10
d)25.9-25.17/-8.80-8.10=25(9-17)/8(-80-10)
=25.(-8)/8(-90)
=5/18
e) 34.5-36/34.13-34
=34(5-32)/34(13-1)
=5-32/13-1=-1/3
KH
20 tháng 4 2020
\(b,\)Để \(\frac{x+3}{x-2}\) là số nguyên thì \(x+3⋮x-2\)
Đặt \(A=\frac{x+3}{x-2}\)
Ta có :\(A=\frac{x-2+5}{x-2}\)
\(A=1+\frac{5}{x-2}\)
Do đó : \(5⋮x-2\) hoặc \(x-2\inƯ\left(5\right)\)
\(\Rightarrow x-2\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow x\in\left\{1;-3;3;7\right\}\)
Vậy .......
TQ
1
21 tháng 7 2017
25.9-25.17/-8.80-.10=25.(9-17)/-8(80+10)=25.(-8)/-8.90=25/90=5/18
48.12-48.15/-3.270-3.30=48.(12-15)/-3(270+30)=48.-3/-3.300=48/300=4/25
thông cảm e mới học lớp 6 nên sai thì thôi còn chỗ -3 hay -8 là vì thay -2 lần trừ ra cộng
HH
1
28 tháng 4 2017
a/\(\frac{3939-101}{3.2929+505}=\frac{39.101-101}{8787+505}=\frac{101.\left(39-1\right)}{87.101+5.101}=\frac{101.38}{101.\left(87+5\right)}=\frac{38}{92}\)
\(=\frac{38}{92}\)
28 tháng 4 2017
b/\(\frac{6.4+6.7}{6.5+12}=\frac{4+1.7}{1.5+2}=\frac{4+7}{5+2}=\frac{11}{7}\)
22 tháng 2 2015
a) \(\frac{25.9-25.17}{-8.80-8.10}=\frac{25.\left(9-17\right)}{-8.\left(80+10\right)}=\frac{25.\left(-8\right)}{-8.90}=\frac{5}{18}\)
b) \(\frac{48.12-48.15}{-3.270-3.30}=\frac{48.\left(12-15\right)}{-3.\left(270+30\right)}=\frac{48.\left(-3\right)}{-3.300}=\frac{4}{25}\)
c) \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{2^5.8}{2^5.\left(25-3\right)}=\frac{2^5.8}{2^5.22}=\frac{4}{11}\)
d) \(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{3^4.\left(5-9\right)}{3^4.14}=\frac{3^4.\left(-4\right)}{3^4.14}=\frac{-2}{7}\)
2 tháng 7 2020
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
a ) \(\frac{-360}{450}\)
TA có : \(\frac{-360}{450}=\frac{-4}{5}\)
b ) \(\frac{-260}{1500}\)
Ta có : \(\frac{-260}{1500}\)= \(\frac{-13}{75}\)