Tính 27 3 - - 8 3 - 125 3
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a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)
b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)
c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)
a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) = \(\dfrac{6}{7}\)
b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)
c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)
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a) ( − 125 ) . ( − 5 ) .8. ( − 2 ) = ( − 125 ) .8. ( − 5 ) . ( − 2 ) = − 1000.10 = − 10000
b) ( − 127 ) . ( 1 − 582 ) − 582.127 = ( − 127 ) . ( − 581 ) − 582.127 = 127.581 − 582.127 = 127 ( 581 − 582 ) = 127. ( − 1 ) = − 127
c) ( 43 − 13 ) . ( − 3 ) + 27. ( − 14 − 16 ) = 30. ( − 3 ) + 27. ( − 30 ) = ( − 30 ) .3 + 27. ( − 30 ) = ( − 30 ) . ( 3 + 27 ) = ( − 30 ) .30 = − 90
d) 125. ( − 61 ) . ( − 2 ) 3 . ( − 1 ) 2 n = 125. ( − 62 ) . ( − 8 ) .1 = 125. ( − 8 ) . ( − 62 ) = − 1000. ( − 62 ) = 62000
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Giải
a) (−8).(−3)3.(+125)(−8).(−3)3.(+125)
= [(−2).(−2).(−2)].[(−3).(−3).(−3)].(5.5.5)[(−2).(−2).(−2)].[(−3).(−3).(−3)].(5.5.5)
= [(−2).(−3).5].[(−2).(−3).5].[(−2).(−3).5][(−2).(−3).5].[(−2).(−3).5].[(−2).(−3).5]
= 30.30.30=30330.30.30=303
b) 27.(−2)3.(−7).(+49)27.(−2)3.(−7).(+49)
= (3.3.3).[(−2).(−2).(−2)].[(−7).(−7).(−7)](3.3.3).[(−2).(−2).(−2)].[(−7).(−7).(−7)]
= [3.(−2).(−7)].[3.(−2).(−7)].[3.(−2).(−7)][3.(−2).(−7)].[3.(−2).(−7)].[3.(−2).(−7)]
= 42.42.42=423
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\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)
\(=6\sqrt{3}-4\sqrt{\dfrac{1}{3}}-4\sqrt{3}-5\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-9\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{9\cdot\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{3}\)
\(=-\sqrt{3}\)
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\(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)
\(=\left(5\sqrt{5}-2\sqrt{3}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+3\sqrt{3}\right)\)
\(=\left(3\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}+2\sqrt{3}\right)\)
\(=\left(3\sqrt{5}\right)^2-\left(2\sqrt{3}\right)^2\)
\(=15-12\)
\(=3\)
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a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
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P/s : mk làm từng phần một
\(\left(\frac{2}{3}\right)^x=\frac{8}{27}=\left(\frac{2}{3}\right)^3\)
=> x = 3
Vậy,........
2)
\(\left(\frac{5}{6}x+\frac{1}{2}\right)^2=\frac{9}{16}=\left(\frac{3}{4}\right)^2\)
\(\frac{5}{6}x+\frac{1}{2}=\frac{3}{4}\)
\(\frac{5}{6}x=\frac{1}{4}\)
\(x=\frac{3}{10}\)
∛27 - ∛-8 - ∛125 = ∛33 - ∛(-2)3 - ∛53
= 3 - (-2) - 5 = 3 + 2 - 5 = 0