Tính giá trị biểu thức:
A = (1/2)^-10 × 8^-3 +(0,2)^-4 × 25^-2 + 81^-1 × (1/3)^-7
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a/ 35 – 2.1111 + 3. 72
= 35 - 2.1 + 3.49
= 35 - 2 + 147
= 33 + 147
= 180
b/ 5.43 + 2.3 – 81. 2 + 7
= 5.64 + 6 - 162 + 7
= 320 + 6 - 162 + 7
= 326 - 162 + 7
= 164 + 7
= 171
c/ 25 + 2.{12 + 2.[3.(5-2)+1]+1}+1
= 32 + 2.{12 + 2.[3.3+1]+1}+1
= 32 + 2.{12 + 2.10+1}+1
= 32 + 2.{12 + 20 + 1}+1
= 32 + 2.{32+1}+1
= 32 + 2.33 + 1
= 32 + 66 + 1
= 98 + 1
= 99
d/ 17.131 + 69.17
= 17 . (131+69)
= 17 . 200
= 3400
e/ 50 – [30 - (6 - 2)2 ]
= 50 - [30 - 42]
= 50 - 14
= 36
f/ 199 +36 + 201 + 184 + 37
= (199+201)+(36+184)+37
= 400 + 220 + 37
= 657
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,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023
=0+0+...+0-1-2023
=-2024
a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)
b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)
Câu 3:
\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)
\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4:
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)
a: \(2\dfrac{3}{5}+1\dfrac{2}{5}\cdot\dfrac{31}{2}\)
\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{31}{2}\)
\(=\dfrac{26}{10}+\dfrac{217}{10}=\dfrac{243}{10}\)
b: \(4\dfrac{3}{4}-3\dfrac{2}{3}:1\dfrac{1}{6}\)
\(=\dfrac{19}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)
\(=\dfrac{19}{4}-\dfrac{11}{3}\cdot\dfrac{6}{7}\)
\(=\dfrac{19}{4}-\dfrac{22}{7}\)
\(=\dfrac{19\cdot7-22\cdot4}{28}=\dfrac{45}{28}\)
`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600