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14 tháng 10

 

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12 tháng 12 2019

a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)

\(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)

=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)

\(\frac{-5}{12}-\frac{8}{5}\)

\(\frac{\left(-25\right)-96}{60}\)

\(\frac{\left(-25\right)+\left(-96\right)}{60}\)

=\(\frac{121}{60}\)

b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)

\(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)

=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)

\(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)

=\(\frac{-29}{24}\)

c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)

     =\(\frac{3}{5}\)

d)\(\frac{5^4.20^4}{25^5.4^5}\)

=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)

=\(\frac{100^4}{100^5}\)

=\(\frac{1}{100}\)

21 tháng 6 2022

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

 

13 tháng 7 2022

a)\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}

2 \cdot x=\dfrac{1}{5}

x=\dfrac{1}{5}: 2

 x=\dfrac{1}{10}
b) \left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}

-1 \dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}
x=2+1 \dfrac{3}{5}

 x=3 \dfrac{3}{5}
c) \left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}

x \cdot \dfrac{3}{7} \cdot \dfrac{1}{7}=\dfrac{-3}{8}

x=\dfrac{-3}{8}: \dfrac{3}{49}
x=\dfrac{-49}{8}=-6 \dfrac{1}{8}
d) \dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)

\dfrac{-4}{7} x+\dfrac{7}{5}=\dfrac{1}{8} \cdot \dfrac{-3}{5}
-\dfrac{4}{7} x=\dfrac{-3}{40}-\dfrac{7}{5} \\ x=\dfrac{-59}{40}: \dfrac{-4}{7}=\dfrac{413}{160}=2 \dfrac{93}{160}
 

22 tháng 8 2019

a, \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)

\(=\left(\frac{1}{4}+\frac{5}{12}\right)-\left(\frac{1}{13}+\frac{7}{8}\right)\)

\(=\frac{2}{3}-\frac{99}{104}\)

\(=-\frac{89}{312}\)

b, \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)

\(=\left(11\frac{3}{13}+5\frac{3}{13}\right)-2\frac{4}{7}\)

\(=\frac{214}{13}-\frac{18}{7}\)

\(=\frac{1264}{91}\)

c, \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)

\(=6\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)

\(=\left(6\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)

\(=2+3\frac{7}{11}\)

\(=5\frac{7}{11}\)

\(=\frac{62}{11}\)

d, \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\cdot0\)

\(=0\)

e, \(-1,5\cdot\left(1+\frac{2}{3}\right)\)

\(=-\frac{3}{2}\cdot\frac{5}{3}\)

\(=-\frac{5}{2}\)

f, Đặt \(A=1^2+2^2+3^2+...+100^2\)

\(=1+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)

\(=1+2\cdot3-2+3\cdot4-3+...+100\cdot101-100\)

\(=\left(2\cdot3+3\cdot4+...+100\cdot101\right)-\left(1+2+3+...+100\right)\)

Đặt B = 2 . 3 + 3 . 4 + ... + 100 . 101 

3B = 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 100 . 101 . ( 102 - 99 )

3B = 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 100 . 101 . 102 - 99 . 100 . 101 

3B = 100 . 101 . 102

B = \(\frac{100\cdot101\cdot102}{3}\)

B = 343400

Thay B vào A. Ta được :

\(A=343400-\left(1+2+3+...+100\right)\)

Thay C = 1 + 2 + 3 + ... + 100

Dãy số 1; 2; 3; ...; 100 có số số hạng là:

( 100 - 1 ) : 1 + 1 = 100 ( số hạng )

Tổng của dãy số đó là :

( 100 + 1 ) . 100 : 2 = 5050

=> C = 5050

Thay C vào A. Ta được :

\(A=343400-5050\)

\(A=338350\)

Vậy A = 338350