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
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\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)
\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)
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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.
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\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
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
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Ta có : \(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{n}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(\dfrac{a+1}{a}\right)^2-2.\dfrac{a+1}{a}.\dfrac{1}{a+1}+\left(\dfrac{1}{a+1}\right)^2}=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=1+\dfrac{1}{a}-\dfrac{1}{a+1}\left(a>0\right)\) Áp dụng điều này vào bài toán trên , ta được :
\(P=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\) \(P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}\)
\(P=98+\dfrac{1}{2}-\dfrac{1}{100}\)
\(P=\dfrac{9849}{100}\)
Với a+b+c =0 (a,b,c \(\ne\) 0) , ta có:
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)
Áp dụng cho từng thừa số của P, ta có:
\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{\left(1+\dfrac{1}{2}+\dfrac{1}{-3}\right)-2\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)}\)
\(=\sqrt{\left(1+\dfrac{1}{2}-\dfrac{1}{3}\right)^2-2.\dfrac{3-2-1}{6}}=\left|1+\dfrac{1}{2}-\dfrac{1}{3}\right|=1+\dfrac{1}{2}-\dfrac{1}{3}\)
Tương tự :\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}=1+\dfrac{1}{3}-\dfrac{1}{4}\)
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=1+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=1+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{149}{100}\)
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Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
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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:
\(B=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2011}-\frac{1}{2012}\)
\(=2011+(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012})\)
\(=2012-\frac{1}{2012}\)
2/ \(\sqrt{4+\sqrt{4+...+\sqrt{4}}}< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{7+\sqrt{4}}}}}=3\)
1/ Ta có:
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\left(\dfrac{n^2+n+1}{n\left(n+1\right)}\right)^2}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=99+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=100-\dfrac{1}{100}=\dfrac{9999}{100}\)
Em cảm ơn ạ