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\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2-2.\sqrt{2}.1+1}}{\sqrt{17-3.2.2.\sqrt{2}}}-\)\(\frac{\sqrt{2+2.\sqrt{2}.1+1}}{\sqrt{17+3.2.2.\sqrt{2}}}\)
\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{17-3.2.\sqrt{4}.\sqrt{2}}}\)\(-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{17+3.2.\sqrt{4}.\sqrt{2}}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{8-2.\sqrt{8}.3+9}}\)\(-\frac{\sqrt{2}+1}{\sqrt{8+2.\sqrt{8}.3+9}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{\left(\sqrt{8}-3\right)^2}}\)\(-\frac{\sqrt{2}+1}{\sqrt{\left(\sqrt{8}+3\right)^2}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{8}-3}\)\(-\frac{\sqrt{2}+1}{\sqrt{8}+3}\)
\(=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{8}+3\right)-\left(\sqrt{2}+1\right)\left(\sqrt{8}-3\right)}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{\sqrt{16}+3\sqrt{2}-\sqrt{8}-3-\sqrt{16}+3\sqrt{2}-\sqrt{8}+3}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{6\sqrt{2}-2\sqrt{8}}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{6\sqrt{2}-2.2.\sqrt{2}}{\left(2\sqrt{2}-3\right)\left(2\sqrt{2}+3\right)}\)
\(=\frac{2\sqrt{2}}{\left(8-9\right)}=\frac{2\sqrt{2}}{-1}=-2\sqrt{2}\)
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)-\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}=\frac{\sqrt{2}+1-\left(\sqrt{2}-1\right)}{9-8}=2\)
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
A= căn (5-2 (căn 5) +1)-căn (5+2 (căn 5) +1)
=căn ((căn 5)-1)^2 -căn ((căn 5)+1)^2
=l (căn 5) -1l - l (căn 5) +1l
=căn 5 -1 -căn 5 -1
=-2
A, biến đổi 6= căn bậc hai của 5 + 1 -> hằng đẳng thức
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