Tính:
a, \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
b, \(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
K
Khách
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TD
5 tháng 10 2017
Bài 68 :
a ) \(\sqrt[3]{27}-\sqrt[3]{8}-\sqrt[3]{125}=3-2-5=-4\)
b ) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=\sqrt[3]{27}-\sqrt[3]{216}=3-6=-3\)
TD
5 tháng 10 2017
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 \(3< \sqrt[3]{123}\)
Y
22 tháng 6 2023
\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
\(=-5\)
\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=\sqrt{5}.\left(5-12+6-4\right)\)
\(=-5\sqrt{5}\)
\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)
\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)
\(=2.3+5-5.2\)
\(=1\)
\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)
\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)
\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)
\(=0\)
\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
22 tháng 6 2023
\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)
\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
L=2*3+5-5*2=5-4=1
N=8căn 2-8căn2-15căn3+15căn 3=0
O=10căn 3-6căn3+3căn3=7căn 3
31 tháng 10 2021
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
18 tháng 9 2019
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
19 tháng 9 2019
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
19 tháng 6 2017
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)\(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
\(=3+2-5\)
\(=0\)
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, \(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
\(=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54.4}\)
\(=\sqrt[3]{27}-\sqrt[3]{216}\)
\(=3-6\)
\(=-3\)