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\(A=\frac{2}{\sqrt{5}-3}-\frac{2}{\sqrt{5}+3}=\frac{2\left(\sqrt{5}+3\right)-2\left(\sqrt{5}-3\right)}{-4}=\frac{2\sqrt{5}+6-2\sqrt{5}+6}{-4}=\frac{12}{-4}=-3\)
Vay ........
a. \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
= \(\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}\)
= \(-2\sqrt{3}\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
= \(\dfrac{\left(\sqrt{15}-\sqrt{5}\right).\left(\sqrt{3}+1\right)}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(2\sqrt{5}+4\right)}{4}\)
=\(\dfrac{\sqrt{45}+\sqrt{15}-\sqrt{15}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).2\left(\sqrt{5}+2\right)}{4}\)
= \(\dfrac{3\sqrt{5}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(\sqrt{5}+2\right)}{2}\)
= \(\dfrac{2\sqrt{5}}{2}+\dfrac{5\sqrt{5}+10-10-4\sqrt{5}}{2}\)
= \(\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
= \(\dfrac{3\sqrt{5}}{2}\)
c. \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
= \(\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right).\left(\sqrt{5}+\sqrt{2}\right)}.\left(\sqrt{2}+1\right)^2\)
= \(\dfrac{2\sqrt{5}}{3}.\left(2+2\sqrt{2}+1\right)\)
= \(\dfrac{2\sqrt{5}}{3}.\left(3+2\sqrt{2}\right)\)
= \(\dfrac{6\sqrt{5}+4\sqrt{10}}{3}\)
d. \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-3\left(\sqrt{3}+2\right)+\dfrac{5\left(3+\sqrt{3}\right)}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(-2\sqrt{3}-5+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{-4\sqrt{3}-10+15+5\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{\sqrt{3}+5}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{1}{2}\)
Nếu đúng cho 1 like nhé!
a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
a) \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2\sqrt{3}+2-\sqrt{3}\)
\(=\left(2\sqrt{3}-\sqrt{3}\right)+2\)
\(=\sqrt{3}+2\)
b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{1+\sqrt{5}}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{1+\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{12}{4}=3\)
c) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{1}=14\)
a: \(=\left(2\sqrt{7}+\sqrt{7}+2\sqrt{14}\right)\cdot\sqrt{7}-\left(51+14\sqrt{2}\right)\)
\(=3\sqrt{7}\cdot\sqrt{7}+2\sqrt{14}\cdot\sqrt{7}-51-14\sqrt{2}\)
\(=21-51=-30\)
b: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{10}-\sqrt{6}}{2}=\dfrac{2\sqrt{10}-\sqrt{6}}{2}\)
c: \(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}+\dfrac{\left(\sqrt{5}-\sqrt{2}\right)^2}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{5}+\sqrt{3}-\sqrt{2}\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
Ta có:
Đáp án: (C)