Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(10+\sqrt{60}+\sqrt{24}+\sqrt{40}=10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}\)
\(=\left(5+2\sqrt{15}+3\right)+2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)+2\)
\(=\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{10+\sqrt{60}+\sqrt{24}+\sqrt{40}}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
Dùng hẳng đẳng thức 3 số:
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
$VT=\sqrt{5+3+2+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}=\sqrt{(\sqrt5+\sqrt3+\sqrt2)^2}=VP(đpcm)$
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
= \(\sqrt{\sqrt{10}\left(\sqrt{10}+\sqrt{6}\right)+\sqrt{4}\left(\sqrt{6}+\sqrt{10}\right)}\)
=\(\sqrt{\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{10}+\sqrt{4}\right)}\)
Lời giải:
\(A=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{10+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2\sqrt{15}}=\sqrt{2+(3+5+2\sqrt{15})+2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{2+(\sqrt{3}+\sqrt{5})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(2B=2.\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2(1+\sqrt{2})\)
Do đó:
\(A-2B=\sqrt{3}+\sqrt{5}-(2+\sqrt{2})>\sqrt{2}+\sqrt{4}-(2+\sqrt{2})=0\)
\(\Rightarrow A>2B\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2+3+5+2\sqrt{2.3}+2\sqrt{2.5}+2\sqrt{3.5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
Bạn tham khảo tại link sau:
Câu hỏi của Vi Huỳnh - Toán lớp 9 | Học trực tuyến
\(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
\(=\sqrt{2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}-\sqrt{2}\right)^2\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)