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\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}=\dfrac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}=\dfrac{4+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}}{2\sqrt{2}}\) \(=\dfrac{4+2\sqrt{7}-2\sqrt{7}}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
ta có:\(\sqrt{45+27\sqrt{2}}\) +\(\sqrt{45-27\sqrt{2}}\) =3(\(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\))
phân thức cần biến đổi trở thành:\(\dfrac{\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}\).3
=\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}=\dfrac{5+\sqrt{7}}{\sqrt{2}}=\dfrac{5\sqrt{2}+\sqrt{14}}{2}\)
lấy phân số vừa tìm được cộng cho \(\dfrac{\sqrt{14}}{2}\) ta được giá trị biểu thức cần tìm là \(\dfrac{5\sqrt{2}}{2}\)
Tại sao kết quả lại bằng, có thể giải thích hộ mình ko\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}\)
a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
Ta có: \(A=\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}\right)^2-\left(\sqrt{5-3\sqrt{2}}\right)^2}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{\left(\sqrt{3+\sqrt{2}}\right)^2-\left(\sqrt{3-\sqrt{2}}\right)^2}\)
\(=\frac{3\left[\left(\sqrt{5+3\sqrt{2}}\right)^2+2\cdot\sqrt{5+3\sqrt{2}}\cdot\sqrt{5-3\sqrt{2}}+\left(\sqrt{5-3\sqrt{2}}\right)^2\right]}{\left|5+3\sqrt{2}\right|-\left|5-3\sqrt{2}\right|}-\frac{\left(\sqrt{3+\sqrt{2}}\right)^2+2\cdot\sqrt{3+\sqrt{2}}\cdot\sqrt{3-\sqrt{2}}+\left(\sqrt{3-\sqrt{2}}\right)^2}{\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|}\)
\(=\frac{3\left(\left|5+3\sqrt{2}\right|+2\sqrt{7}+\left|5-3\sqrt{2}\right|\right)}{5+3\sqrt{2}-\left(5-3\sqrt{2}\right)}-\frac{\left|3+\sqrt{2}\right|+2\cdot\sqrt{7}+\left|3-\sqrt{2}\right|}{3+\sqrt{2}-\left(3-\sqrt{2}\right)}\)
\(=\frac{3\left(5+3\sqrt{2}+2\sqrt{7}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\frac{3+\sqrt{2}+2\sqrt{7}+3-\sqrt{2}}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\frac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\frac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\frac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\frac{3\left(6+2\sqrt{7}\right)}{6\sqrt{2}}\)
\(=\frac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}\)
\(=\frac{12}{6\sqrt{2}}=\sqrt{2}\)
Vậy: \(A=\sqrt{2}\)
Lời giải:
a) Đặt \(\sqrt{3+\sqrt{2}}=a; \sqrt{3-\sqrt{2}}=b\Rightarrow \left\{\begin{matrix}
a^2+b^2=6\\
a^2-b^2=2\sqrt{2}\\
ab=\sqrt{(3+\sqrt{2})(3-\sqrt{2})}=\sqrt{7}\end{matrix}\right.\)
\(A=\frac{a+b}{a-b}=\frac{(a+b)^2}{(a+b)(a-b)}=\frac{(a+b)^2}{a^2-b^2}=\frac{a^2+b^2+2ab}{a^2-b^2}\)
\(=\frac{6+2\sqrt{7}}{2\sqrt{2}}=\frac{3+\sqrt{7}}{\sqrt{2}}\)
b)
Đặt \(\sqrt{5+3\sqrt{2}}=a; \sqrt{5-3\sqrt{2}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=10\\ a^2-b^2=6\sqrt{2}\\ ab=\sqrt{(5+3\sqrt{2})(5-3\sqrt{2})}=\sqrt{25-(3\sqrt{2})^2}=\sqrt{7}\end{matrix}\right.\)
\(B=\frac{\sqrt{9(5+3\sqrt{2})}+\sqrt{9(5-3\sqrt{2})}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}=\frac{3(a+b)}{a-b}\)
\(=\frac{3(a+b)^2}{a^2-b^2}=\frac{3(a^2+b^2+2ab)}{a^2-b^2}=\frac{3(10+2\sqrt{7})}{6\sqrt{2}}\)
\(=\frac{5+\sqrt{7}}{\sqrt{2}}\)
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: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
\(\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{\left(\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{5+3\sqrt{2}-\left(5-3\sqrt{2}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{3+\sqrt{2}-\left(3-\sqrt{2}\right)}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{6\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)
\(=\frac{10+2\sqrt{7}-6-2\sqrt{7}}{2\sqrt{2}}=\sqrt{2}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{\sqrt{9\left(5+3\sqrt{2}\right)}+\sqrt{9\left(5-3\sqrt{2}\right)}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{3+\sqrt{2}+2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+3-\sqrt{2}}{2\sqrt{2}}\)
\(=\dfrac{3\left[5+3\sqrt{2}+5-3\sqrt{2}+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\right]}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}=\dfrac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}=\dfrac{12}{6\sqrt{2}}\)
\(=\sqrt{2}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\sqrt{5+3\sqrt{2}}+3\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)}-\dfrac{\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}}{\sqrt{6+3\sqrt{2}}-\sqrt{6-4\sqrt{2}}}\\ =\dfrac{3\left(5+3\sqrt{2}+2\sqrt{25-18}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{\sqrt{4+2+4\sqrt{2}}+\sqrt{4+2-4\sqrt{2}}}{\sqrt{4+2+4\sqrt{2}}-\sqrt{4+2-4\sqrt{2}}}\\ =\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{2+\sqrt{2}+2-\sqrt{2}}{2+\sqrt{2}-2+\sqrt{2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{4}{2\sqrt{2}}=\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)