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\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\frac{2-\sqrt{3}}{\sqrt{2}+\sqrt{2}-\sqrt{3}}\)
= \(\frac{2+\sqrt{3}}{2\sqrt{2}+\sqrt{3}}+\frac{2-\sqrt{3}}{2\sqrt{2}-\sqrt{3}}\)
= \(\frac{\left(2+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}\)
= \(\frac{4\sqrt{2}-2\sqrt{3}+2\sqrt{6}-3+4\sqrt{2}+2\sqrt{3}-2\sqrt{6}-3}{\left(2\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}\)
= \(\frac{8\sqrt{2}-6}{-5}\)
Chúc bạn học tốt !!!
GIẢI
\(M=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
\(=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)
\(=\frac{\sqrt[2]{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\left(\sqrt{5}+1\right)}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)
\(=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{5+1-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
Chúc bạn học tốt !!!
Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\sqrt{\dfrac{\left(1+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{2}}\)
= \(\dfrac{1+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\)
<=> A = 1
=\(\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)+\(\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\sqrt{2}-\sqrt{3}}\)+.....+\(\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right).\left(\sqrt{99}-\sqrt{100}\right)}\)
=\(\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{99}\)
=\(-1+\sqrt{100}\)
=9
\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\) \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)