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b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)
Ta chứng minh công thức:
\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
Với n\(\in N\)* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)
a) Áp dụng (*) vào T
\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)
\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)
Vậy n=24.
Bài 1
\(1+\frac{1}{a^2}+\frac{1}{(a+1)^2}=(1+\frac{1}{a})^2-\frac{2}{a}+\frac{1}{(a+1)^2}\)
\(=(\frac{a+1}{a})^2-2.\frac{a+1}{a}.\frac{1}{a+1}+(\frac{1}{a+1})^2=(\frac{a+1}{a}-\frac{1}{a+1})^2\)
\(=(1+\frac{1}{a}-\frac{1}{a+1})^2\)
$\Rightarrow A=|1+\frac{1}{a}-\frac{1}{a+1}|=1+\frac{1}{a}-\frac{1}{a+1}$ với $a>0$
Bài 2:
Áp dụng kết quả bài 1 thì:
\(B=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2011}-\frac{1}{2012}\)
\(=2011+(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012})\)
\(=2012-\frac{1}{2012}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
Chứng minh công thức:
\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a^2+2a+1\right)+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+a^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\dfrac{\left(a^2\right)^2+2a^2a+2a^2+2a+a^2+1}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}}\)
\(=\dfrac{a^2+a+1}{a\left(a+1\right)}\)
\(=\dfrac{a\left(a+1\right)+1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a}-\dfrac{1}{a+1}\)
Áp dụng công thức ta có:
\(C=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1999^2}+\dfrac{1}{2000^2}}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=2000-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
hình như \(\dfrac{1}{2}\) không bị trừ bn ạ