\(T=\dfrac{\sqrt{1}-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}\)
\(\Rightarrow T=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)
\(\Rightarrow T=\sqrt{4}-\sqrt{1}\)
\(\Rightarrow T=2-1=1\)
Tính :
\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...+\dfrac{1}{\sqrt{2017}-\sqrt{2018}}\)
1.Thu gọn
A=\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2021}}\)
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
Rút gọn:
A = \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(a,2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(b,\sqrt{32}-\sqrt{50}+\sqrt{18}\)
\(c,3\sqrt{3}+4\sqrt{2}-5\sqrt{27}\)
\(d,\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
e,\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
a,Rút gọn K
b,Tính K khi x=\(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
c,Tìm x để \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\)≥1
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. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
Rút gọn
A=\(\sqrt{13+4\sqrt{10}}\)
B= \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
C= \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
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