Tính
a ) 27 3 − − 8 3 − 125 3 b ) 135 3 5 3 − 54 3 ⋅ 4 3
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Bài 68 :
a ) \(\sqrt[3]{27}-\sqrt[3]{8}-\sqrt[3]{125}=3-2-5=-4\)
b ) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=\sqrt[3]{27}-\sqrt[3]{216}=3-6=-3\)
Bài 69 :
a ) Ta có : \(\left\{{}\begin{matrix}3^3=27\\\left(\sqrt[3]{123}\right)^3=123\end{matrix}\right.\)
Vì 27 < 123 nên suy ra \(3< \sqrt[3]{123}\)
Vậy \(3< \sqrt[3]{123}\)
LG a
3√27−3√−8−3√125273−−83−1253
Phương pháp giải:
Tính từng căn bậc ba rồi thực hiện phép tính
Lời giải chi tiết:
3√27−3√−8−3√125=3√33−3√(−2)3−3√53273−−83−1253=333−(−2)33−533
=3−(−2)−5=3−(−2)−5
=3+2−5=0=3+2−5=0.
LG b
3√1353√5−3√54.3√4135353−543.43
Phương pháp giải:
Sử dụng các công thức:
3√a.b=3√a.3√ba.b3=a3.b3.
3√ab=3√a3√bab3=a3b3, với b≠0b≠0.
Lời giải chi tiết:
3√1353√5−3√54.3√4=3√27.53√5−3√54.4135353−543.43=27.5353−54.43
=3√5.3√273√5−3√216=53.27353−2163
=3√27−3√216=273−2163
=3√33−3√63=333−633
=3−6=−3=3−6=−3.
a)\(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
\(=3+2-5\)
\(=0\)
b)\(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
\(=\sqrt[3]{\frac{153}{5}}-\sqrt[3]{54.4}\)
\(=\sqrt[3]{\frac{153}{5}}-6\)
Theo mình câu b như vậy
pham trung thanh câu b bn làm thiếu hay sao ý? Theo tôi, cả bài làm như thế này.
Giải:
a, \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
\(=\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{12}=3+2-5\)
\(=0\)
b, \(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
\(=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54.4}\)
\(=\sqrt[3]{27}-\sqrt[3]{216}\)
\(=3-6\)
\(=-3\)
a) \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
= \(\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
= \(-16\sqrt{3}\)
b) \(\left(a.\sqrt{\dfrac{a}{b}}+2\sqrt{ab}+b.\sqrt{\dfrac{b}{a}}\right)\sqrt{\dfrac{a}{b}}\)
= \(\dfrac{a^2}{b}+2a+b\) = \(\dfrac{a^2+\left(2a+b\right)b}{b}\) = \(\dfrac{a^2+2ab+b^2}{b}\) = \(\dfrac{\left(a+b\right)^2}{b}\)
c) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\) = \(3+2-5=0\)
d) \(3+\sqrt{18}+\sqrt{3}+\sqrt{8}\) = \(3+3\sqrt{2}+\sqrt{3}+2\sqrt{2}\)
= \(3+\sqrt{3}+5\sqrt{2}\)
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)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)
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
a ) 27 3 − − 8 5 − 125 3 = 33 3 − ( − 2 ) 3 3 − 53 3 = 3 − ( − 2 ) − 5 = 3 + 2 − 5 = 0 b ) 135 3 5 3 − 54 3 ⋅ 4 3 = 135 5 3 − 54.4 3 = 27 3 − 216 3 = 33 3 − 63 3 = 3 − 6 = − 3