Giải:
Ta có:
\(-7\sqrt{3}=-\sqrt{7^2.3}=-\sqrt{147}\)
\(-2\sqrt{10}=-\sqrt{2^2.10}=-\sqrt{40}\)
Vì \(\sqrt{147}>\sqrt{40}\)
\(\Leftrightarrow-\sqrt{147}< -\sqrt{40}\)
Vậy ...
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
So sánh \(\sqrt{2015}+\sqrt{2018}\) và \(\sqrt{2016}+\sqrt{2017}\)
bài 1 : rút gọn
a)\(\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
b)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
c)\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
bài 2
a)\(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}\)
b)\(\frac{\sqrt{6}-5\sqrt{3}}{2\sqrt{2}-10}\)
c) \(\frac{7-2\sqrt{10}}{5-\sqrt{10}}\)
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn
a,\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b,\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
c,\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\text{So sánh }\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\text{ với }2\) 2
Làm hộ mình câu c nha
Cho \(H=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\).
a) Rút gọn H
b) Chứng minh \(H\ge0\)
c) So sánh H với \(\sqrt{H}\)
So sánh x và y trong các TH sau: \(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}};y=\sqrt{2017}+\sqrt{2018}\)
