Tính
a) \(\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)
b) \(\frac{18.12-48.15}{-3.270-3.30}\)
c) \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
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24 tháng 6 2022
a: \(\dfrac{25\cdot9-25\cdot17}{-8\cdot80-8\cdot10}=\dfrac{25\cdot\left(-8\right)}{\left(-8\right)\cdot\left(80+10\right)}=\dfrac{25}{90}=\dfrac{5}{18}=\dfrac{250}{900}\)
\(\dfrac{48\cdot12-48\cdot15}{\left(-3\right)\cdot\left(270+30\right)}=\dfrac{48}{300}=\dfrac{8}{50}=\dfrac{144}{900}\)
\(\dfrac{2^5\cdot3}{600}=\dfrac{96}{600}=\dfrac{144}{900}\)
b: \(\dfrac{2^5\cdot7+25}{25\cdot5^2-2^5\cdot3}=\dfrac{32\cdot7+25}{25\cdot25-32\cdot3}=\dfrac{249}{529}=\dfrac{1743}{7\cdot529}\)
\(\dfrac{3^4\cdot5-3^6}{3^4\cdot13+3^4}=\dfrac{3^4\left(5-9\right)}{3^4\left(13+1\right)}=\dfrac{-4}{14}=\dfrac{-2}{7}=\dfrac{-1058}{7\cdot529}\)
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) \(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{3^4.\left(5-9\right)}{3^4.14}=\frac{3^4.\left(-4\right)}{3^4.14}=\frac{-2}{7}\)
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
\(a)\frac{{5{\rm{x}} + 10}}{{25{{\rm{x}}^2} + 50}} = \frac{{5\left( {x + 2} \right)}}{{25\left( {{x^2} + 2} \right)}} = \frac{{x + 2}}{{5\left( {{x^2} + 2} \right)}}\)
\(b)\frac{{45{\rm{x}}\left( {3 - x} \right)}}{{15{\rm{x}}{{\left( {x - 3} \right)}^2}}} = \frac{{3\left( {3 - x} \right)}}{{{{\left( {x - 3} \right)}^2}}}\)
\(c)\frac{{{{\left( {{x^2} - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {{x^3} + 1} \right)}} = \frac{{\left( {{x^2} - 1} \right)\left( {{x^2} - 1} \right)}}{{\left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - x + 1}}\)
HQ
Hà Quang Minh
Giáo viên
13 tháng 1
a) Ta có: \(P = \frac{{x + 1}}{{{x^2} - 1}} = \frac{{x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{x - 1}}\)
Suy ra: \(Q = \frac{1}{{x - 1}}\)
b) Thay x = 11 vào P ta được: \(P = \frac{{11 + 1}}{{{{11}^2} - 1}} = \frac{1}{{10}}\)
Thay x = 11 vào Q ta được: \(Q = \frac{1}{{11 - 1}} = \frac{1}{{10}}\)
Hai kết quả P = Q tại x = 11
a) (-25).(75-45)-75.(45-25)=(-25).30-75.20
\(a,\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)
\(=\left(-25\right).20-75.20\)
\(=20.\left(-25-75\right)\)
\(=20.\left(-100\right)\)
\(=-2000\)
\(b,\frac{18.12-48.15}{-3.270-3.30}\)
\(=\frac{18.12-12.4.15}{-3.270-3.30}\)
\(=\frac{12.\left(18-60\right)}{-3.\left(270+30\right)}\)
\(=\frac{12.\left(-42\right)}{-3.300}\)
\(=\frac{14}{25}\)
\(c,\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(=\frac{2^5.\left(7+1\right)}{2^5\left(25-3\right)}\)
\(=\frac{2^5.8}{2^5.22}\)
\(=\frac{8}{22}\)
\(=\frac{4}{11}\)