A=\(\dfrac{\sqrt[3]{384}}{\sqrt[3]{3}}+3\sqrt[3]{-54}+\sqrt[3]{432}\)
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12 tháng 6 2023
a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)
b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)
c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)
d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)
1 tháng 9 2021
b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)
\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)
=3-6
=-3
11 tháng 7 2021
a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)
\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)
\(=-5\sqrt{7}\)
b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)
\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)
\(=\sqrt{5}\cdot3\sqrt{5}=15\)
c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)
=-1
e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{4\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-3\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)
\(=0\)
f) Ta có: \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3\sqrt{2}+2\sqrt{3}-2\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
17 tháng 12 2021
\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
21 tháng 8 2018
a , \(A=\sqrt[3]{1000}-\sqrt[3]{-54}-\sqrt[3]{128}=10+3,77976315-5,0396842=8,74007895\)
b , tương tự
9 tháng 10 2021
\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)
\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
AT
\(\left(\dfrac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\dfrac{\sqrt{54}}{3}\right).\dfrac{a}{\sqrt{6}}=-1\)
1
BN
19 tháng 8 2017
\(\left(\dfrac{3\sqrt{2}+6}{\sqrt{12}+2}-\dfrac{\sqrt{54}}{3}\right).\dfrac{a}{3}=-1\)
\(\Leftrightarrow\left[\dfrac{\sqrt{6}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}-\dfrac{3\sqrt{6}}{3}\right].\dfrac{a}{\sqrt{6}}=-1\)
\(\Leftrightarrow\left(\dfrac{\sqrt{6}}{2}-\sqrt{6}\right).\dfrac{a}{\sqrt{6}}=-1\)
\(\Leftrightarrow\sqrt{6}\left(\dfrac{1}{2}-1\right).\dfrac{a}{\sqrt{6}}=-1\)
\(\Leftrightarrow-\dfrac{1}{2}.a=-1\)
\(\Leftrightarrow a=2\)
Vậy a=2
16 tháng 9 2023
\(A=\dfrac{\sqrt[4]{7\sqrt[3]{54}+15\sqrt[3]{128}}}{\sqrt[3]{\sqrt[4]{32}}+\sqrt[3]{9\sqrt[4]{162}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{7\sqrt[3]{3^3.2}+15\sqrt[3]{4^3.2}}}{\sqrt[3]{\sqrt[4]{2^4.2}}+\sqrt[3]{9\sqrt[4]{3^4.2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{7.3\sqrt[3]{2}+15.4\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{9.3\sqrt[4]{2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{21\sqrt[3]{2}+60\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{3^3\sqrt[4]{2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{81\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}=\dfrac{3\sqrt[4]{\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}\)
\(\Leftrightarrow A=\dfrac{3}{\sqrt[3]{2}+3}\)
\(A=\sqrt[3]{\dfrac{384}{3}}+3\cdot\left(-3\right)\cdot\sqrt[3]{2}+6\sqrt[3]{2}\)
\(=4\sqrt[3]{2}-9\sqrt[3]{2}+6\sqrt[3]{2}\)
\(=\sqrt[3]{2}\)