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28 tháng 4 2020

Nhận xét x > 0 

=> \(x^3=10+3x\left(\sqrt[3]{5^2-4.13}\right)\)

<=> \(x^3=10-9x\)

<=> \(x^3+9x-10=0\)

<=> \(\left(x-1\right)\left(x^2+x+10\right)=0\)

<=> \(x-1=0\) vì x > 0 

<=> x = 1 thử lại thỏa mãn

Vậy x = 1 

21 tháng 8 2019

\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

<=>\(x^3=5+2\sqrt{13}+3.\sqrt[3]{5+2\sqrt{13}}.\sqrt[3]{5-2\sqrt{13}}\left(\sqrt[3]{5-2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

<=> \(x^3=10+3\sqrt[3]{5^2-\left(2\sqrt{13}\right)^2}.x\)

<=> \(x^3=10+3\sqrt[3]{-27}.x=10-9x\)

<=> x3+9x-10=0

<=> x3-x2+x2-x+10x-10=0

<=>\(x^2\left(x-1\right)+x\left(x-1\right)+10\left(x-1\right)=0\)

<=> \(\left(x^2+x+10\right)\left(x-1\right)=0\)

<=> \(\left(x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{39}{4}\right)\left(x-1\right)=0\)

<=> \(\left[\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\right]\left(x-1\right)=0\)

=> x-1=0 (vì \(\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\))

<=> x=1

24 tháng 6 2021

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)

\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

24 tháng 6 2021

Khiếp CTV kìa sợ quá ;-;

\(=\sqrt[3]{25-52}=-3\)

14 tháng 7 2019

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

\(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

==================================================

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

===========================================================

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

================================================================

30 tháng 6 2018

\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)

\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)

\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)

1 tháng 7 2018

Câu a

\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)

1 tháng 7 2017

a,\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}=5+3\sqrt{2}\)

1 tháng 7 2017

b, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)

\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)

\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)

\(\Leftrightarrow2\sqrt{3}\)

a) Ta có: \(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)

\(=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}-1-\sqrt{5}-1=-2\)

b) Ta có: \(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{2}+2\sqrt{2}+1}\)

\(=\sqrt{14+32\sqrt{2}}\)

c) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)

\(=\sqrt{6+2\sqrt{5}-2\sqrt{3}-1}\)

\(=\sqrt{5+2\sqrt{5}-2\sqrt{3}}\)