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Áp dụng \(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\) ta có:
\(x=\sqrt{1+\dfrac{1}{\left(\dfrac{1}{999}\right)^2}+\dfrac{1}{\left(\dfrac{1}{999}+1\right)^2}}+\dfrac{999}{1000}=1+\dfrac{1}{\dfrac{1}{999}}-\dfrac{1}{\dfrac{1}{999}+1}+\dfrac{999}{1000}=1+999-\dfrac{999}{1000}+\dfrac{999}{1000}=1000\)
đầu bài phải là: cmr: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)chì bn???
Giải:
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}-2.\left(\frac{b+a-a-b}{ab.\left(a+b\right)}\right)}\)
\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}-2.\left(\frac{1}{a.\left(a+b\right)}+\frac{1}{b.\left(a+b\right)}-\frac{1}{ab}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
=> đpcm
AD: \(\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}=\left|1+999-\frac{999}{1000}\right|+\frac{999}{1000}\)
\(=1000-\frac{999}{1000}+\frac{999}{1000}=1000\)
\(VT=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\left(\frac{2}{ab}-\frac{2}{a\left(a+b\right)}-\frac{2}{b\left(a+b\right)}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\frac{2\left(a+b\right)-2b-2a}{ab\left(a+b\right)}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|=VP\)
Áp dụng tính M: \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
\(M=999.\sqrt{\frac{1}{999^2}+\frac{1}{1^2}+\frac{1}{\left(999+1\right)^2}}+\frac{999}{1000}\)
\(M=999.\left(\frac{1}{1}+\frac{1}{999}-\frac{1}{1000}\right)+\frac{999}{1000}\)
\(M=999+1-\frac{999}{1000}+\frac{999}{1000}=1000\)
Vậy M=1000.
- Gỉa sử \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}=\left(\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\right)^2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}+\frac{2}{ab}-\frac{2}{b\left(a+b\right)}-\frac{2}{a\left(a+b\right)}\)
=> \(\frac{2}{ab}-\frac{2}{b\left(a+b\right)}-\frac{2}{a\left(a+b\right)}=0\)
=> \(\frac{a+b}{ab\left(a+b\right)}-\frac{a}{ab\left(a+b\right)}-\frac{b}{ab\left(a+b\right)}=0\)
=> \(\frac{a+b-a-b}{ab\left(a+b\right)}=\frac{0}{ab\left(a+b\right)}=0\) (Luôn đúng )
Vậy ....
- Áp dụng : \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
=> \(M=\sqrt{1+999^2+\frac{999^2}{\left(1+999\right)^2}}+\frac{999}{1000}\) ( với \(a=1,b=999\) )
=> \(M=1+999-\frac{999}{1000}+\frac{999}{1000}=1000\)
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
\(P=\sqrt{1+999^2+\dfrac{999^2}{1000^2}+\dfrac{999}{1000}}\)
\(\Leftrightarrow\)\(\sqrt{\dfrac{1999}{1000}+999^2+\dfrac{999^2}{1000^2}}\)
???