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Xét với n là số tự nhiên không nhỏ hơn 1
Ta có : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng điều trên ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
\(=1-\frac{1}{\sqrt{2002}}< 1-\frac{1}{\sqrt{2025}}=1-\frac{1}{45}=\frac{44}{45}\)
ta chứng minh công thức tổng quát sau
\(\frac{1}{\left[n+1\right]\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left[n+1\right]}\left[\sqrt{n+1}+\sqrt{n}\right]}\)
=\(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}\left[n+1-n\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left[n+1\right]}}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
ta có \(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
........
\(\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}=\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=> \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+..+\frac{1}{2002\sqrt{2001}+2001\sqrt{2002}}\)
=\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2001}}-\frac{1}{\sqrt{2002}}\)
=\(1-\frac{1}{\sqrt{2002}}< \frac{44}{45}\)
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Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x^2-x+2001}=a\\\sqrt[3]{3x^2-7x+2002}=b\\\sqrt[3]{6x-2003}=c\end{matrix}\right.\)
\(\Rightarrow a^3-b^3-c^3=2002\) từ đây ta có:
\(a-b-c=\sqrt[3]{a^3-b^3-c^3}\)
\(\Leftrightarrow\left(a-b-c\right)^3=\sqrt[3]{a^3-b^3-c^3}\)
\(\Leftrightarrow\left(a-c\right)\left(a-b\right)\left(b+c\right)=0\)
Tự làm nốt nhé
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sao bạn lại có chữ hiệp sĩ ở bên cạnh tên vậy?
sao vậy bạn
k mk nha
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(x+2004-2004+4)/2000+(x-2004+2004+3)/2001=(x-2004+2004+2)/2002+(x-2004+2004+1)/2003
hay (x+2004)/2000-1+(x+2004)/2001-1=(x+2004)/2002-1+(x+2004)/2003-1
Hay (x+2004)(1/2000+1/2001)=(x+2004)(1/2002+1/2003)
Hay (x+2004)(1/2000+1/2001-1/2002-1/2003)=0
hay x+2004=0
Hay x=-2004
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Đặt \(\sqrt[3]{3x^2-x+2001}=a;-\sqrt[3]{3x^2-7x+2002}=b;-\sqrt[3]{6x-2003}=c\)
Thì ta có được hệ: \(\hept{\begin{cases}a+b+c=\sqrt[3]{2002}\\a^3+b^3+c^3=2002\end{cases}}\)
\(\Leftrightarrow\left(a+b+c\right)^3=a^3+b^3+c^3\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=a^3+b^3+c^3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Với a = - b thì
\(\sqrt[3]{3x^2-x+2001}=\sqrt[3]{3x^2-7x+2002}\)
\(\Leftrightarrow3x^2-x+2001=3x^2-7x+2002\)
\(\Leftrightarrow6x=1\)
\(\Leftrightarrow x=\frac{1}{6}\)
Tương tự cho 2 trường hợp còn lại
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\(x^4+\sqrt{x^2+2002}=2002\)
Đặt \(\sqrt{x^2+2002}=a^2>0\)
\(\Rightarrow\hept{\begin{cases}x^4+a^2=2002\left(1\right)\\a^4-x^2=2002\left(2\right)\end{cases}}\)
Lấy (1) - (2) ta được
\(x^4-a^4+x^2+a^2=0\)
\(\Leftrightarrow\left(x^2+a^2\right)\left(x^2-a^2+1\right)=0\)
\(\Leftrightarrow x^2+1=a^2=\sqrt{x^2+2002}\)
\(\Leftrightarrow x^4+2x^2+1=x^2+2002\)
\(\Leftrightarrow x^4+x^2-2001=0\)
Tới đây thì đơn giản rồi
\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
\(\Leftrightarrow\left(x^2+3x+1\right)^2=\left(x+3\right)^2\left(x^2+1\right)\)
\(\Leftrightarrow x^2=8\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{8}\\x=-\sqrt{8}\end{cases}}\)
\(A=\left(\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{2002-1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{2002}{2002!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{2001!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\frac{1}{1!}-\frac{1}{2002!}+\frac{1}{2002!}=1\)