\(\dfrac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}< \sqrt{3}+\sqrt{4}=\dfrac{1}{\sqrt{4}-\sqrt{3}}\)
\(\dfrac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}< \sqrt{3}+\sqrt{4}=\dfrac{1}{\sqrt{4}-\sqrt{3}}\)
1.Thu gọn
A=\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2021}}\)
Tính :
\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...+\dfrac{1}{\sqrt{2017}-\sqrt{2018}}\)
tính tổng \(T=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}\)
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}vs10\) so sánh
1) Tính giá trị biểu thức C=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\) 2) Chứng minh rằng với mọi số nguyên dương n ta đêu có \(\sqrt{4+\sqrt{4+\sqrt{4+\sqrt{4+...+\sqrt{4}}}}}\) < 3 ( n căn bậc 4) Mọi người giúp em với ạ
\(a,2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(b,\sqrt{32}-\sqrt{50}+\sqrt{18}\)
\(c,3\sqrt{3}+4\sqrt{2}-5\sqrt{27}\)
\(d,\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
e,\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
\(D=\sqrt[]{\dfrac{3\sqrt{ }3-4}{2\sqrt{3+1}}+\sqrt[]{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}}\)
Tính:
a) \(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5\)
b) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2\)
c) \(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)
d) \(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
e) \(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
C/m các đẳng thức sau:
a) \(\sqrt{21-6\sqrt{6}}\) + \(\sqrt{9+2\sqrt{18}}\) - \(2\sqrt{6+3\sqrt{3}}\) = 0
b) \(\dfrac{1}{\sqrt{25}+\sqrt{24}}\) + \(\dfrac{1}{\sqrt{24}+\sqrt{23}}\) \(\dfrac{1}{\sqrt{23}+\sqrt{22}}\) +...+ \(\dfrac{1}{\sqrt{2}+\sqrt{1}}\) = 4
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\) = \(\sqrt{2}\) - 1
Mn giúp mk với !!!