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Sửa đề: A=7/12-9/20+11/30-13/42+15/56
A=1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8
=1/3+1/8=11/24
a) (-12). (7 - 72) - 25. (55 - 43)
= (-12). (- 65) - 25. 12
= 12. 65 – 12. 25
= 12. (65 - 25)
= 12. 40
= 480
b) (39 - 19) : (- 2) + (34 - 22). 5
= 20 : (- 2) + 12. 5
= - 10 + 60
= 60 - 10
= 50.
a) Với a = 3,05 thì ta có:
\(a\times2,46\) \(=3,05\times2,46=7,503\)
b) Với \(a=\dfrac{15}{8}\) thì ta có:
\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\) \(=\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\) \(=\left(\dfrac{10}{12}+\dfrac{7}{12}\right)\times\dfrac{8}{15}\) \(=\dfrac{17}{12}\times\dfrac{8}{15}\) \(=\dfrac{34}{45}\)
a) Thay a=3,05 vào 2,46a, ta được:
\(2.46\cdot3.05=7.503\)
b) Thay \(a=\dfrac{15}{8}\) vào biểu thức \(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\), ta được:
\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\)
\(=\dfrac{17}{12}\cdot\dfrac{8}{15}=\dfrac{34}{45}\)
\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)
|(5/8-5/14)+(3/8-9/14)|:4/7
=|(5/8+3/8)+(-5/14-9/14)|:4/7
=|1+(-1)|:4/7
=0
\(a,=\dfrac{5}{6}+\dfrac{6}{7}\\ =\dfrac{35}{42}+\dfrac{36}{42}=\dfrac{71}{42}\\ b,=\dfrac{5}{6}+\dfrac{4}{9}\\ =\dfrac{15}{18}+\dfrac{8}{18}=\dfrac{23}{18}\)
a: \(=2\sqrt{5}-2\sqrt{5}+9\sqrt{5}-30\sqrt{5}=-21\sqrt{5}\)
b: \(=2\sqrt{7}-6\sqrt{7}-\dfrac{3}{4}\sqrt{7}-8\sqrt{7}=-\dfrac{51}{4}\sqrt{7}\)
a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{14}{5}\)
b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{9}{5}+\dfrac{5}{5}\)
\(=\dfrac{14}{5}\)
\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
