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C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)
C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)
C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)
C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)
C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)
D = \(-6\sqrt{5}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
a: \(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{2+\sqrt{3}}{2}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{2-\sqrt{3}}{2}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\dfrac{\sqrt{3}+1}{2}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\dfrac{\sqrt{3}-1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}\cdot\dfrac{2}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2}\cdot\dfrac{2}{2-\sqrt{3}+1}\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(=\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
\(=\dfrac{6}{6}=1\)
\(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}\)
\(=\sqrt{6}-\sqrt{2}\)
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
bai 1
(n+1)√n=√n^3+√n>2√(n^3.n)=2n^2>2(n^2-1)=2(n-1)(n+1)
1/[(n+1)√n]<1/[2(n-1)(n+1)]=1/4.[2/(n-1)(n+1)]
A=..
n =1 yes
n>1
A<1+1/4[2/1.3+2/3.5+..+2/(n-1)(n+1)
A<1+1/4[ 2-1/(n+1)]<1+1/2<2=>dpcm
Câu a : Ta có :
\(\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{1-\sqrt{2}}{1-2}=\dfrac{1-\sqrt{2}}{-1}=-1+\sqrt{2}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{2}-\sqrt{3}}{2-3}=\dfrac{\sqrt{2}-\sqrt{3}}{-1}=-\sqrt{2}+\sqrt{3}\)
.....................
\(\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}=\dfrac{\sqrt{n^2-1}-\sqrt{n^2}}{\left(\sqrt{n^2-1}+\sqrt{n^2}\right)\left(\sqrt{n^2-1}-\sqrt{n^2}\right)}=\dfrac{\sqrt{n^2-1}-\sqrt{n^2}}{-1}=-\sqrt{n^2-1}+\sqrt{n^2}\)
Thay vào ta được :
\(S=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...........-\sqrt{n^2-1}+\sqrt{n^2}\)
\(=-1+\sqrt{n^2}\)
Câu b:
Đặt biểu thức đã cho là $A$
Ta có:
\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)
\(\Leftrightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)
\(\Leftrightarrow A> \frac{1}{2}(n-1)\) (áp dụng cách tính toán phần a)
Lại có:
\(A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{1+\sqrt{2}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\right)+....+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-3}+\sqrt{n^2-2}}+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)
\(\Leftrightarrow A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)
\(\Leftrightarrow A< \frac{\sqrt{n^2-1}}{2}\) (áp dụng cách tính toán của phần a)
Vậy \(\frac{\sqrt{n^2-1}}{2}> A> \frac{n-1}{2}\) hay \(\sqrt{t(t+1)}> A> t\) (đặt \(n=2t+1\) )
Mà \(\sqrt{t(t+1)}< \sqrt{(t+1)(t+1)}=t+1\)
Do đó: \(t+1> A> t\)
\(\Rightarrow \lfloor{A}\rfloor=t=\frac{n}{2}\)