Tính:
a. 3. (-7/2 + 2/-3 +-3/5)
b. 3/5 + -2/3 - (-5/2)- 13/30
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Lời giải:
a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)
b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)
c.
\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)
\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)
d.
\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)
\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)
\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)
a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)
b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)
d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=5+3\sqrt{2}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
2:
a: \(2^2\cdot3-4=4\cdot3-4=12-4=8\)
b: \(16-2^3\cdot2=16-2^4=16-16=0\)
c: \(4^2-4\cdot2=16-8=8\)
d: \(3^3-2\cdot3^2=27-2\cdot9=27-18=9\)
e: \(7^2-9\cdot2^2=49-9\cdot4=49-36=13\)
f: \(2^2\cdot3+4^2=4\cdot3+16=12+16=28\)
Bài 1:
a: \(13+21\cdot5-\left(198:11-8\right)\)
\(=13+105-18+8\)
=21+87
=108
b: \(272:16-5+4\cdot\left(30-5-255:17\right)\)
\(=17-5+4\cdot\left(30-5-15\right)\)
\(=12+4\cdot10=12+40=52\)
c: \(15\cdot24-14\cdot5\cdot\left(145:5-27\right)\)
\(=360-70\left(29-27\right)\)
=360-140
=220
d: \(18\cdot3-18\cdot2+3\cdot\left(\dfrac{51}{17}\right)\)
\(=18\left(3-2\right)+3\cdot3\)
=18+9
=27
e: \(64+115+36-25\cdot8\cdot6\cdot2^2\cdot3+4^2\)
\(=100+115-200\cdot6\cdot4\cdot3+16\)
\(=231-14400=-14169\)
f: \(15\cdot8-\left(17-30+83\right)-144:6\)
\(=120-24-\left(100-30\right)\)
=96-70
=26
2:
a: =>2/3:x=1,4-2,4=-1
=>x=-2/3
b: =>x/5=25/30-19/30=6/30=1/5
=>x=1
3:
Số học sinh giỏi là 40*1/4=10 bạn
Số học sinh khá là 30*3/5=18 bạn
Số học sinh TB là 30-18=12 bạn
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82
a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
a: \(-15-\left(-13+30\right)\)
\(=-15+13-30\)
\(=\left(-15-30\right)+13\)
=-45+13
=-32
b: \(225-150:\left(30+3^2\cdot5\right)\)
\(=225-150:\left(30+9\cdot5\right)\)
\(=225-150:75\)
=225-2
=223
c: \(18\cdot64+18\cdot36-1200\)
\(=18\left(64+36\right)-1200\)
\(=18\cdot100-12\cdot100\)
\(=6\cdot100=600\)
d: \(80-\left[130-\left(12-4\right)^2\right]\)
\(=80-130+\left(12-4\right)^2\)
\(=-50+8^2=64-50=14\)
Tính:
a, 8.( -9/12 )
b, ( -14/5 ) . (-10)
c, 9. ( 4/-3 )
d, -15. ( -5/6 )
e, 13. ( -2/7 )
f, 7/26. ( -13 )
\(a,8.\left(\dfrac{-9}{12}\right)=\dfrac{8.\left(-9\right)}{12}=\dfrac{-72}{12}=-6\\ b,\left(\dfrac{-14}{5}\right).\left(-10\right)=\dfrac{\left(-14\right).\left(-10\right)}{5}=\dfrac{140}{5}=28\\ c,9.\left(\dfrac{4}{-3}\right)=\dfrac{9.\left(-4\right)}{3}=\dfrac{-36}{3}=-12\\ d,-15.\left(\dfrac{-5}{6}\right)=\dfrac{\left(-15\right).\left(-5\right)}{6}=\dfrac{75}{6}\\ e,13.\left(\dfrac{-2}{7}\right)=\dfrac{13.\left(-2\right)}{7}=\dfrac{-26}{7}\\ f,\dfrac{7}{26}.\left(-13\right)=\dfrac{7.\left(-13\right)}{26}=\dfrac{-91}{26}\)
a, 8.( -9/12 )=-6
b, ( -14/5 ) . (-10)=28
c, 9. ( 4/-3 )=-12
d, -15. ( -5/6 )=25/2
e, 13. ( -2/7 )-36/7
f, 7/26. ( -13 )=-7/2
`a)\sqrt{9-4sqrt5}-sqrt5`
`=sqrt{5-2.2sqrt5+4}-sqrt5`
`=sqrt{(sqrt5-2)^2}-sqrt5`
`=|\sqrt5-2|-sqrt5`
`=sqrt5-2-sqrt5=-2`
`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`
`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`
`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`
`=|2-sqrt3|+|sqrt3-1|`
`=2-sqrt3+sqrt3-1=1`
`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`
`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`
`=sqrtx+7`
`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`
`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`
`=sqrt3+1-2sqrt3-1=-sqrt3`
`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)
lớp 5 đã học âm đâu NGUYỄN THÚY AN
a)\(3.\left(\frac{-7}{2}+\frac{2}{-3}+\frac{-3}{5}\right)\)
=\(3.\left(\frac{-25}{6}+\frac{-3}{5}\right)\)
=\(3.\frac{-143}{30}\)
=\(\frac{-143}{10}\)
=-14,3