Tính
a ) 17 5 − 3 5 ; b ) − 3 4 − 4 5 ; c ) − 5 6 − − 5 9 d ) 6 12 − 7 − 21 ; e ) 3 2 − 1 ; f ) 2 − − 4 5 .
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a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}\)
=\(\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)\)
=\(\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
b) \(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{17}+\dfrac{-2}{5}\)
=\(\left(\dfrac{5}{17}-\dfrac{2}{17}\right)-\left(\dfrac{9}{15}+\dfrac{2}{5}\right)\)
=\(\dfrac{3}{17}-1=\dfrac{-14}{17}\)
\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)
\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)
\(A=2-1+\frac{1}{3}\)
\(A=\frac{4}{3}\)
\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)
\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=18+\frac{-5}{13}+\frac{-18}{35}\)
\(B=\frac{7781}{455}\)
\(a,157.17-157.7=157\left(17-7\right)=1570\)
\(b,5.\left(-3+2\right)-7.\left(5-4\right)=-5-7=-12\)
\(c,17.\left(-84\right)+17\left(-16\right)=17\left[\left(-84\right)+\left(-16\right)\right]=-1700\)
\(d,-145.\left(13-57\right)+57.\left(10-145\right)=6380-7695=-1315\)
\(e,199.\left(15-17\right)-199.\left(-17+5\right)=199.10=1990\)
b,\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-2}{3}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-4}{6}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).\(\frac{1}{6}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{18}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{6}\).\(\frac{1}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{12}\)
=\(\frac{8}{12}\)+\(\frac{1}{12}\)
=\(\frac{9}{12}\)=\(\frac{3}{4}\)
a)=18.17-18.7
=18.(17-7)=18.10=180
b)=54-(6.17+6.9)=54-102-54=(54-54)-102=0-102=-102
c)=(33.17-33.5)-(17.33-17.5)=33.17-33.5-17.33+17.5=(33.17-17.33)-5.(33-17)=0-80=-80
Ta có: \(a^3=\left(\sqrt[3]{3+\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\right)^3\)
\(=3+\sqrt{17}+3-\sqrt{17}+3\sqrt[3]{\left(3+\sqrt{17}\right)\left(3-\sqrt{17}\right)}\left(\sqrt[3]{3+\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\right)\)
(\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) )
\(=6+3\sqrt[3]{-8}.a=6-6a\)
\(\Rightarrow a^3+6a-6=0\Rightarrow a^3+6a-5=1\)
\(\Rightarrow A=1^{2019}=1\)
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
a ) 14 5 b ) − 31 20 c ) − 5 18
d ) 5 6 . e ) 1 2 . f ) 14 5