CMR: A=B
a) A=\(\sqrt{10+\sqrt{24+\sqrt{40+\sqrt{60}}}}\)
B= \(\sqrt{5}\)+\(\sqrt{3}\)+\(\sqrt{2}\)
b) A=\(\sqrt{3+\sqrt{5}}\)
B= \(\frac{\sqrt{5}+1}{\sqrt{2}}\)
CMR:\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2+3+5+2\sqrt{2.3}+2\sqrt{2.5}+2\sqrt{3.5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
Chứng minh các hằng đẳng thức sau:
a) \(y\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
b) \(\sqrt{6+\sqrt{24+\sqrt{12}+\sqrt{8}}}-\sqrt{3}=\sqrt{2}+1\)
Chứng minh các hằng đẳng thức:
a) \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
b) \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1\)
Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)
Cho: A = \(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
B = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
So sánh A và 2B
Bạn tham khảo tại link sau:
Câu hỏi của Vi Huỳnh - Toán lớp 9 | Học trực tuyến
Tính :
a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=8\)
\(A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(A^2=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\) (do \(A>0\))
\(C=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{3}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}-\frac{\sqrt{3}-\sqrt{2}}{3}\)
\(=\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{3}=\frac{\sqrt{2}}{2}\)
Tính :
a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)
a)=\(\sqrt{3-\sqrt{5}}\).\(\sqrt{3+\sqrt{5}}\).\(\sqrt{2}\)(\(\sqrt{5}\)-\(1\))\(\sqrt{3+\sqrt{5}}\)=2\(\sqrt{2}\) \(\sqrt{\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\) .\(\sqrt{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\)\(\sqrt{8}\) =8
b)A2=8+2 căn[\(\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)\)]=8+2\(\sqrt{6-2\sqrt{5}}\)=8+2(\(\sqrt{5}\)-1)=6+2\(\sqrt{5}\)=(\(\sqrt{5}+1\))2 =>A=\(\sqrt{5}\)+1
c)C=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)
Rút gọn các biểu thức sau:
a.\(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
b.\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
c.\(\sqrt{8+\sqrt{40}+\sqrt{20}+\sqrt{8}}\)
d.\(\sqrt{10+\sqrt{24}+\sqrt{20}+\sqrt{8}}\)
d.\(\sqrt{10+\sqrt{24}-\sqrt{40}-\sqrt{60}}\)
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
a/ \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}\)
\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)
\(=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=2\sqrt{3+\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\)
\(=\sqrt{2}+\sqrt{6}\)
Rút gọn biểu thức
a) A= \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b) B= \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
c) C= \(\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\)
Lời giải:
a)
\(\frac{2A}{\sqrt{2}}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{3+1+2\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3}}}+\frac{3+1-2\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3}}}\)
\(=\frac{(\sqrt{3}+1)^2}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{(\sqrt{3}-1)^2}{2-\sqrt{(\sqrt{3}-1)^2}}=\frac{(\sqrt{3}+1)^2}{2+\sqrt{3}+1}+\frac{(\sqrt{3}-1)^2}{2-(\sqrt{3}-1)}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{3}(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)^2}{\sqrt{3}(\sqrt{3}-1)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}=2\)
$\Rightarrow A=\sqrt{2}$
b)
\(B=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}=\sqrt{(8+2\sqrt{15})+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
c)
\(C=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}=\frac{\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+2\sqrt{x-1}+1}}{\sqrt{(x-2)^2}}\)
\(=\frac{\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}}{|x-2|}=\frac{|\sqrt{x-1}-1|+|\sqrt{x-1}+1|}{|x-2|}\)
Thực hiện phép tính
a) (\(2\sqrt{3}-\sqrt{2}\))2+\(2\sqrt{24}\)
b) \(\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}\)
\(a,\left(2\sqrt{3}-\sqrt{2}\right)^2+2\sqrt{24}=\left[\left(2\sqrt{3}\right)^2-2.2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\right]+2\sqrt{24}\\ =\left[12-4\sqrt{6}+2\right]+2\sqrt{24}=14-4\sqrt{6}+4\sqrt{6}=14\\ b,\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}=3\sqrt{5}.\sqrt{5}-2\sqrt{3}.\sqrt{3}+3\sqrt{5}.2\sqrt{3}-\sqrt{3}.\sqrt{5}-\sqrt{60}\\ =15-6+6\sqrt{15}-\sqrt{15}-\sqrt{2^2.15}\\ =9+3\sqrt{15}\)