tính
a)1 + 3/4 b)4/5 -3/8 c)1 : 2/3 d)2/5 x 5/2
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`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`-7/25 + (-8)/25`
`= (-7 - 8)/25`
`= -15/25`
`= -3/5`
`b)`
`6/13 + (-15)/39`
`= 18/39 + (-15)/39`
`= (18 - 15)/39`
`= 3/39`
`= 1/13`
`c)`
`5/7 + 4/(-14)`
`= 10/14 + (-4)/14`
`= (10 - 4)/14`
`= 6/14`
`= 3/7`
`d)`
`-8/18 + (-15)/27`
`= -4/9 + (-5)/9`
`= (-4-5)/9`
`= -9/9 = -1`
`2,`
`a)`
`3/5 + (-7)/4`
`= 12/20 + (-35)/20`
`= (12 - 35)/20`
`=-23/20`
`b)`
`(-2) + (-5)/8`
`= (-16)/8 + (-5)/8`
`= (-16 - 5)/8`
`= -21/8`
`c)`
`1/8 + (-5)/9`
`= 9/72 + (-40)/72`
`= (9-40)/72`
`= -31/72`
`d)`
`6/13 + (-14)/39`
`= 18/39 + (-14)/39`
`= (18 - 14)/39`
`= 4/39`
`e)`
`(-18)/24 + 15/21`
`= (-3)/4 + 5/7`
`= (-21)/28 + 20/28`
`= (-21 + 20)/28`
`= -1/28`
a) 3/7 + 3/4 - 1/2 =33/28-1/2=19/28
b) 6/11 + 2/9 x 12 =6/11+8/3=106/33
c) 5/7 - 8/5 : 4 =5/7-2/5=11/35
a) 3/7 + 3/4 - 1/2 = \(\dfrac{12}{28}+\dfrac{21}{28}-\dfrac{14}{28}=\dfrac{19}{28}\)
b) 6/11 + 2/9 x 12 = \(\dfrac{6}{11}+\dfrac{8}{3}=\dfrac{18}{33}+\dfrac{88}{33}=\dfrac{106}{33}\)
c) 5/7 - 8/5 : 4 = \(\dfrac{5}{7}-\dfrac{2}{5}=\dfrac{25}{35}-\dfrac{14}{35}=\dfrac{11}{35}\)
\(d,\dfrac{6}{5}-3:\dfrac{15}{4}=\dfrac{6}{5}-3\times\dfrac{4}{15}=\dfrac{6}{5}-\dfrac{4}{5}=\dfrac{2}{5}\)
\(b,\dfrac{2}{5}+\dfrac{4}{5}:4=\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{1}{4}=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)
\(a,\dfrac{5}{8}+\dfrac{4}{3}=\dfrac{15}{24}+\dfrac{32}{24}=\dfrac{47}{24}\)
\(a,\dfrac{4}{7}+\dfrac{7}{2}\\ =\dfrac{8}{14}+\dfrac{49}{14}\\ =\dfrac{8+49}{14}\\ =\dfrac{57}{14}\)
\(b,\dfrac{5}{8}\times\dfrac{3}{2}\\ =\dfrac{5\times3}{8\times2}\\ =\dfrac{15}{16}\)
\(c,\dfrac{3}{2}\times\dfrac{5}{6}-\dfrac{2}{3}\\ =\dfrac{3\times5}{2\times6}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{8}{12}\\ =\dfrac{15-8}{12}\\ =\dfrac{7}{12}\)
\(d,\dfrac{13}{15}+\dfrac{2}{5}:\dfrac{3}{4}\\ =\dfrac{13}{15}+\dfrac{2}{5}\times\dfrac{4}{3}\\ =\dfrac{13}{15}+\dfrac{2\times4}{5\times3}\\ =\dfrac{13}{15}+\dfrac{8}{15}\\ =\dfrac{13+8}{15}\\ =\dfrac{21}{15}\\ =\dfrac{7}{5}\)
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
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)
a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)
b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)
c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)
d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)
a)1 + 3/4=7/4
b)4/5 -3/8=17/40
c)1 : 2/3=3/2
d)2/5 x 5/2=1