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tham khảo: Câu hỏi của Lê Thị Ngọc Duyên - Toán lớp 9 | Học trực tuyến
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 = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
2) a) \(VT=\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\dfrac{1}{\sqrt{n+1}.\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}\)
\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}=VP\left(đpcm\right)\)
b) Áp dụng công thức câu a), ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
câu 1a
9972=9972-9+9
=(997-3)(997+3)+9
=1000.994+9=994000+9
=994009
b) bạn trục mẫu đi nha dựa vào hằng đẳng thức a^2 -b^2=(a-b)(a+b)
rồi bạn tính nói chung mẫu bằng -1
tính cái trên tử kết quả là 4
c) bạn dựa vào câu b .\(\dfrac{1}{\sqrt{3}}=\dfrac{2}{2\sqrt{3}}>\dfrac{2}{\sqrt{3}+\sqrt{4}}\)
từ đó suy ra B > 2A vậy B>8
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
Lời giải:
\(S_n=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+..+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Để \(S_n\in\mathbb{Z}\Rightarrow \sqrt{n+1}-1\in\mathbb{Z}\Rightarrow \sqrt{n+1}\in\mathbb{Z}\)
Đặt \(\sqrt{n+1}=t\in\mathbb{N}>1\) do \(n>0\)
\(\Rightarrow n+1=t^2\Rightarrow t^2\leq 101\) do \(n\leq 100\)
\(\Rightarrow 0< t\leq \sqrt{101}\)
Mà \(t\in\mathbb{N}^*\Rightarrow t\in\left\{1;2;3;4;5;6;7;8;9;10\right\}\)
\(\Rightarrow n=t^2-1\in\left\{3; 8; 15; 24;35;48;63;80;99\right\}\)
ĐK: \(n\le\dfrac{625}{4}\le156\) (vì \(n\in Z\) )
Đặt \(a=\sqrt{\dfrac{25}{2}+\sqrt{\dfrac{625}{4}-n}}+\sqrt{\dfrac{25}{2}-\sqrt{\dfrac{625}{4}-n}}\) \(\left(a\ge0,a\in Z\right)\)
\(\Rightarrow a^2=25+2\sqrt{\dfrac{625}{4}-\dfrac{625}{4}+n}\)
\(\Rightarrow a^2=25+2\sqrt{n}\) (1)
Để \(a\in Z\Rightarrow a^2\in Z\Rightarrow\sqrt{n}\in Z^+\)
Vì \(2\sqrt{n}⋮2\) mà 25 không chia hết cho 2
\(\Rightarrow a^2\) không chia hết cho 2
\(\Rightarrow\) a không chia hết cho 2
Đặt \(a=2k+1\left(k>0,k\in Z\right)\)
\(\left(1\right)\Rightarrow\left(2k+1\right)^2=25+2\sqrt{n}\)
\(\Rightarrow2\sqrt{n}=4k^2+4k-24\)
\(\Rightarrow\sqrt{n}=2k^2+2k-12\)
Vì \(\sqrt{n}\ge0\Rightarrow2k^2+2k-12\ge0\)
\(\Rightarrow\left(k+3\right)\left(k-2\right)\ge0\)
Vì \(k>0\Rightarrow k\ge2\) (2)
Mặt khác: \(n\le156\Rightarrow\sqrt{n}\le\sqrt{156}\) mà \(\sqrt{n}\in Z\)
\(\Rightarrow\sqrt{n}\le12\Rightarrow2k^2+2k-12\le12\)
\(\Rightarrow\left(k-3\right)\left(k+4\right)\le0\)
Vì \(k>0\Rightarrow0< k\le3\) (3)
Từ (2) và (3)\(\Rightarrow\left[{}\begin{matrix}k=2\\k=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}n=0\\n=144\end{matrix}\right.\) (t/m)
Vậy n=0, n=144
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