Tính
\(P=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
Những câu hỏi liên quan
Tính \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
Tính: \(Q=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
Đặt 2015 = a Ta có :
\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a^2+2a+1+1\right)}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a+1\right)^2+a^4+2a^3+2a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
= \(\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2015+1=2016\)
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RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
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Giúp mình với
Tính:
\(P=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
Đề viết sai nha bạn phải là \(-\frac{2015^2}{2016^2}\)
\(=\sqrt{1+2015^2-\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
\(=\sqrt{\left(1+2015-\frac{2015}{2016}\right)^2}+\frac{2015}{2016}\)
\(=1+2015-\frac{2015}{2016}+\frac{2015}{2016}\)
\(=2016\)
tick cho mình nha
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giải phương trình :\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)
Rút gọn \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}+\frac{2015}{2016}}\)
\(x+2015\frac{1}{2}=\sqrt{1+2015^2+\frac{2015^2}{20162}}+\frac{2015}{2016}\)
Rút gọn :
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}.}\)
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
=\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)
=\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
áp dụng vào biểu thức ta có\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
=\(1-\frac{1}{\sqrt{2016}}\)
đến đây cậu tự giải nốt nhé
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bạn coi thử sách VHB đi hình như có đấy
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Chứng minh rằng:\(\frac{43}{44}\le\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\le\frac{44}{45}\)