Question

Giúp mình b4 câu c;d
và những bài còn lại vs ạ

Answer 1

Bài 4:

c: \(\left(2\sqrt{2}-\sqrt{3}\right)^2+4\sqrt{6}\)

\(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2-2\cdot2\sqrt{2}\cdot\sqrt{3}+4\sqrt{6}\)

\(=8+3-4\sqrt{6}+4\sqrt{6}=8+3=11\)

d: \(\left(3\sqrt{3}+\sqrt{5}\right)^2-6\sqrt{15}\)

\(=\left(3\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\cdot3\sqrt{3}\cdot\sqrt{5}-6\sqrt{15}\)

\(=27+5+6\sqrt{15}-6\sqrt{15}=32\)

Bài 5:

a: \(\dfrac{5+\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

b: \(\dfrac{2}{2+\sqrt{3}}+\dfrac{2}{2-\sqrt{3}}\)

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

\(=\dfrac{4-2\sqrt{3}+4+2\sqrt{3}}{4-3}=\dfrac{8}{1}=8\)

c: \(\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{2}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\dfrac{2\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\sqrt{5}+\sqrt{3}+\sqrt{7}+\sqrt{5}=2\sqrt{5}+\sqrt{3}+\sqrt{7}\)

d: \(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{6}-2}{\sqrt{3}-\sqrt{2}}\)

\(=\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}\)

\(=\sqrt{7}+\sqrt{2}\)

Bài 6:

a: \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{-1}=-\sqrt{3}\)

b: \(\sqrt{\dfrac{3}{20}}+\sqrt{\dfrac{1}{60}}-2\sqrt{\dfrac{1}{15}}\)

\(=\sqrt{\dfrac{15}{100}}+\sqrt{\dfrac{15}{900}}-\dfrac{2\sqrt{15}}{15}\)

\(=\dfrac{1}{10}\sqrt{15}+\dfrac{1}{30}\sqrt{15}-\dfrac{2}{15}\sqrt{15}\)

\(=\sqrt{15}\left(\dfrac{1}{10}+\dfrac{1}{30}-\dfrac{2}{15}\right)\)

\(=\sqrt{15}\left(\dfrac{3}{30}+\dfrac{1}{30}-\dfrac{4}{30}\right)=0\)