\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|2\sqrt{5}-3\right|-\left|\sqrt{5}-2\right|=2\sqrt{5}-3-\sqrt{5}+2=\sqrt{5}-1\)

b)\(=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c)\(=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-3-\sqrt{5}+2\)
\(=\sqrt{5}-1\)

b,\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)

c,\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(3-2\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=3-2\sqrt{5}-\sqrt{5}+2=5-3\sqrt{5}\)

\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

minh văn nguyễn

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

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

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

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

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

\(\Rightarrow A^3=5\sqrt{2}-7-3\sqrt[3]{5\sqrt{2}-7}^2.\sqrt[3]{5\sqrt{2}+7}+3\sqrt[3]{5\sqrt{2}-7}.\sqrt[3]{5\sqrt{2}+7}^2-5\sqrt{2}-7=-14-3.\sqrt[3]{\left(5\sqrt{2}-7\right)\left(5\sqrt{2}+7\right)}\left[\sqrt[3]{5\sqrt{2}-7}-\sqrt[3]{5\sqrt{2}+7}\right]=-14-3\sqrt[3]{1}.A=-14-3A\)

\(\Rightarrow A^3=-14-3A\Leftrightarrow A^3+3A+14=0\Leftrightarrow\left(A+2\right)\left(A^2-2A+7\right)=0\Leftrightarrow\left[{}\begin{matrix}A=-2\\A^2-2A+7>0\left(loại\right)\end{matrix}\right.\)

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

Bài 3:

a.

\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)

\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

b.

\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)

\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)

a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

\(=4-2\sqrt{4-3}=4-2=2\)

\(\Rightarrow A=-\sqrt{2}\)

b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)

\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)

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

\(\Rightarrow B=\sqrt{2}\)

 \(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)

\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)

Vậy...

~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?

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...