Tính
\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(\sqrt{\sqrt{11}+1}.\sqrt{\sqrt{11}-1}+\sqrt{10}\)
Tính Q=\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
Tính :\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
ai làm đúng mình tick
Tính giá trị biểu thức
A = \(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
B = \(\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
rút gọn A =\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
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: \(A=\sqrt{2}\)
rút gọn các căn thức sau
B=\(\frac{\sqrt{5-\sqrt{3}}-\sqrt{5+\sqrt{3}}}{\sqrt{5-\sqrt{22}}}+\sqrt{27+10\sqrt{2}}\)C= \(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)D=\(\frac{1}{\sqrt{2}-\sqrt{3}}.\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)A= \(\frac{1}{\sqrt{3+2\sqrt{2}}}+\frac{1}{\sqrt{5+2\sqrt{6}}}+\frac{1}{\sqrt{7+2\sqrt{12}}}+....+\frac{1}{\sqrt{199+2\sqrt{9900}}}\)Tính:
a) \(\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3}-\sqrt{2}}\)
b)\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}\)
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 \(\sqrt{5+3\sqrt{2}}=a; \sqrt{5-3\sqrt{2}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=10\\ a^2-b^2=6\sqrt{2}\\ ab=\sqrt{(5+3\sqrt{2})(5-3\sqrt{2})}=\sqrt{25-(3\sqrt{2})^2}=\sqrt{7}\end{matrix}\right.\)
\(B=\frac{\sqrt{9(5+3\sqrt{2})}+\sqrt{9(5-3\sqrt{2})}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}=\frac{3(a+b)}{a-b}\)
\(=\frac{3(a+b)^2}{a^2-b^2}=\frac{3(a^2+b^2+2ab)}{a^2-b^2}=\frac{3(10+2\sqrt{7})}{6\sqrt{2}}\)
\(=\frac{5+\sqrt{7}}{\sqrt{2}}\)
Thu gọn biểu thức:
\(A=\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(B=\sqrt{\left(1-\sqrt{2020}\right)^2}.\sqrt{2021+2\sqrt{2020}}\)
\(C=\sqrt{\sqrt{3}-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Tính
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(\sqrt{\sqrt{11}+1}.\sqrt{\sqrt{11}-1}+\sqrt{10}\)
\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)