thuc hien phep tinh:
a,\(1\dfrac{1}{2}\). \(2\dfrac{1}{3}\)+ \(1\dfrac{1}{3}\). \(\dfrac{1}{2}\)
b,\(\dfrac{1}{9}\).\(\dfrac{2}{145}\)\(-4\dfrac{1}{3}\). \(\dfrac{2}{145}\)+\(\dfrac{2}{145}\)
K
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HH
18 tháng 6 2018
Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.2\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{292}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}.146+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{5684}{9}\right)\)
\(=-\dfrac{392}{45}\)
Vậy ...
HH
19 tháng 6 2018
Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{1}{6}.25=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{29}{9}\right)\)
\(=-\dfrac{2}{45}\)
Vậy ...
PN
7 tháng 10 2017
a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left(\dfrac{-5}{9}+\dfrac{-13}{9}\right)\)
\(=\dfrac{3}{11}.\left(-2\right)\)
\(=\dfrac{-6}{11}\)
b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)
\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)
\(=\dfrac{77}{9}-\dfrac{9}{5}\)
\(=\dfrac{385}{45}-\dfrac{81}{45}\)
\(=\dfrac{304}{45}\)
c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{696}{261}\)
\(=-\dfrac{692}{261}\)
d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=0+0+0+4-1-1-1\)
\(=4-3\)
\(=1\)
27 tháng 5 2022
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
14 tháng 12 2021
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
ND
Nguyễn Đức Trí
VIP
14 tháng 7 2023
a) \(=\dfrac{254x\left(400-1\right)-145}{254+\left(400-1\right)x253}=\dfrac{254x400-254-145}{254+253x400-253}\)
\(=\dfrac{101600-399}{101200+1}=\dfrac{101211}{101201}=\dfrac{101201+10}{101201}=1+\dfrac{10}{101201}\)
b) \(=\dfrac{5392+\left(600+1\right)x5391}{5392x\left(600+1\right)-69}=\dfrac{5392+600x5391+5391}{5392x600+5392-69}\)
\(=\dfrac{10783+3234600}{3235200+5323}=\dfrac{\text{3245383}}{\text{3240523}}=\dfrac{3240523+60}{3240523}=1+\dfrac{60}{3240523}\)
c) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)
PT
4 tháng 12 2017
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
4 tháng 12 2017
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
13 tháng 6 2022
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
a,9 phần 2
b)\(\frac{1}{9}.\frac{2}{145}-4\frac{1}{3}.\frac{2}{145}+\frac{2}{145}\)
\(=\frac{2}{145}.\left(\frac{1}{9}-\frac{13}{3}+1\right)\)
\(=\frac{2}{145}.\left(-\frac{29}{9}\right)\)
\(=\frac{-2}{45}\)