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\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)
\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)
a)
\((\sqrt2- \sqrt3).(\sqrt2+\sqrt3)\)
=\(\sqrt2.\sqrt2 + \sqrt2.\sqrt3-\sqrt3.\sqrt2+\sqrt3.\sqrt3\)
=\(1.1+1.\sqrt3-\sqrt3.1+\sqrt3.\sqrt3\)
=1+0+3=4
\(a,\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2=2-3=-1\)
\(b,-\left(\sqrt{2}\right)^4+\left(\sqrt{3}\right)^6=-\left(\sqrt{2}^2\right)^2+\left(\sqrt{3}^2\right)^3=-2^2+3^3=-4+27=23\)
\(c,A=\frac{1}{1-\frac{1}{1-2^{-4}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}=\frac{1}{1-\frac{1}{1-\frac{1}{16}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=\frac{1}{1-\frac{1}{\frac{15}{16}}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}\)
\(=\frac{1}{1-\frac{16}{15}}+\frac{1}{1+\frac{2}{3}}=\frac{1}{-\frac{1}{15}}+\frac{1}{\frac{5}{3}}=-15+\frac{3}{5}=-14,4\)
\(d,B=9+99+...+99...9=\left(10-1\right)+\left(100-1\right)+...+\left(100...0-1\right)\)
\(=\left(10+100+...+100...0\right)-\left(1+1+...+1\right)=11...10-50=11...1060\)(có 48 chữ số 1)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
Sau khi ib với Hoàng Nguyễn thì đề bài như sau
Tìm \(n\inℕ\)biết
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
ĐKXĐ: n > 1
Ta đi c/m bài toán tổng quát
\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a}-\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\sqrt{a}-\sqrt{a-1}}{a-a+1}\)
\(=\sqrt{a}-\sqrt{a-1}\)
Áp dụng vào bài toán đc
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
\(\Leftrightarrow\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n}-\sqrt{n-1}=11\)
\(\Leftrightarrow\sqrt{n-1}-1=11\)
\(\Leftrightarrow\sqrt{n-1}=12\)
\(\Leftrightarrow n-1=144\)
\(\Leftrightarrow n=145\left(TmĐKXĐ\right)\)
Vậy n = 145