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1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{4-3}=1\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}.\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
\(A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(B=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\)
\(C=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(7-5\right)=2\sqrt{9}\)
\(D=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(E=\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{5^2-\sqrt{5}^2}=\dfrac{60}{20}=3\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
câu b:
(\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\))^2
\(=\left(5+2\sqrt{6}\right)-\left(5-2\sqrt{6}\right)\)\(-2\sqrt{5+2\sqrt{6}}\sqrt{5-2\sqrt{6}}\)
\(=4\sqrt{6}-2\sqrt{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}\)
\(=4\sqrt{6}-2\sqrt{5^2-\left(2\sqrt{6}\right)^2}\)
\(=4\sqrt{6}-2\sqrt{25-24}=4\sqrt{6}-2\)
mấy câu khác tương tự
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)