Tính
a) 1/2 + 1/3 + 1/4
b) 6/5 + 4/3 + 9/7
c) 90/108 - 40/72
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\(a,\dfrac{2}{3}-\dfrac{5}{6}+\dfrac{3}{4}\)
\(=(\dfrac{2}{3}+\dfrac{3}{4})-\dfrac{5}{6}\)
\(=\dfrac{7}{12}\)
\(b,\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{-4}{13}.\dfrac{6}{7}\)
\(=\dfrac{6}{7}.(\dfrac{8}{13}-\dfrac{4}{13})+(\dfrac{6}{13}.\dfrac{9}{7})\)
\(=(\dfrac{6}{7}.\dfrac{4}{13})+\dfrac{54}{91}\)
\(=\dfrac{24}{91}+\dfrac{54}{91}\)
\(=\dfrac{78}{91}=\dfrac{6}{7}\)
\(c,19.(-26)\)
\(=-494\)
a) `2/3 - 5/6 +3/4`
`=4/6-5/6+3/4`
`=-1/6+3/4`
`=-4/24+18/24`
`=14/24=7/12`
b) `6/7 .8/13 +6/13 .9/7 -4/13 .6/7`
`=6/7 .(8/13-4/13) + 6/13 .9/7`
`=6/7 . 4/13 + 6/13 . 9/7`
`= 6/13 .4/7 +6/13 .9/7`
`=6/13 . (4/7+9/7)`
`=6/13 . 13/7`
`=6/7`
c) `19.(-26)`
`=(1-20).(20+6)`
`=20+6-400-120`
`=-494`
a: =7+5/11-2-3/7-3-5/11
=2-3/7=11/7
b: =-3/5(5/7+3/7+6/7)
=-3/5*2=-6/5
c: =1/3(4/5+6/5)-4/3
=2/3-4/3=-2/3
d: =5/9(7/13+13-3/13)
=5/9*165/13=275/39
a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
a \(\dfrac{53}{84}\)
b \(\dfrac{20}{63}\)
c \(\dfrac{1}{60}\)
a: \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{4}=\dfrac{56}{84}+\dfrac{60}{84}-\dfrac{63}{84}=\dfrac{53}{84}\)
b: \(=\dfrac{3}{7}-\dfrac{1}{9}=\dfrac{27}{63}-\dfrac{7}{63}=\dfrac{20}{63}\)
c: \(=\dfrac{2}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{3}=\dfrac{2}{8}\cdot\dfrac{1}{15}=\dfrac{1}{60}\)
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
a:=5/4-3/4=2/4=1/2
b: =2/3-2/5=10/15-6/15=4/15
c: =10/12-9/12=1/12
d: =4/3-1/4=16/12-3/12=13/12
\(a,\dfrac{20}{16}-\dfrac{3}{4}=\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\ b,\dfrac{30}{45}-\dfrac{2}{5}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ c,\dfrac{10}{12}-\dfrac{3}{4}=\dfrac{5}{6}-\dfrac{3}{4}=\dfrac{10}{12}-\dfrac{9}{12}=\dfrac{1}{12}\\ d,\dfrac{12}{9}-\dfrac{1}{4}=\dfrac{4}{3}-\dfrac{1}{4}=\dfrac{16}{12}-\dfrac{3}{12}=\dfrac{13}{12}\)
A) 13/12
B) 401/105
C) 5/13
a)\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{5}{6}+\frac{1}{4}\)
\(=\frac{13}{12}\)
b)\(\frac{6}{5}+\frac{4}{3}+\frac{9}{7}\)
\(=\frac{38}{15}+\frac{9}{7}\)
\(=\frac{401}{105}\)
c)\(\frac{90}{108}-\frac{40}{72}\)
\(=\frac{5}{6}+\frac{5}{9}\)
\(=\frac{25}{18}\)
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