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binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
Xét hạng tổng quát:
\(\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{\sqrt{n}-\sqrt{n-1}}{n-n+1}=\sqrt{n}-\sqrt{n-1}\)
Áp dụng vào bài, ta có:
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(=\sqrt{n}-1\)
B = \(\dfrac{3}{5}+\dfrac{3}{5^2}+\dfrac{3}{5^3}+...+\dfrac{3}{5^{2016}}\)
=> 5B = \(3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^{2015}}\)
=> 4B = \(3-\dfrac{3}{5^{2016}}\)
=> B = \(\dfrac{3-\dfrac{3}{5^{2016}}}{4}\)
@Nguyễn Đình Dũng có thể đưa ra kết quả chính xác được không?
\(\Rightarrow B=\frac{\sqrt{b}\left(\sqrt{ab}-b\right)-\sqrt{a}\left(a-\sqrt{ab}\right)}{\left(a-\sqrt{ab}\right)\left(\sqrt{ab}-b\right)}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{b\sqrt{a}-\sqrt{b}^3-\sqrt{a}^3+a\sqrt{b}}{a\sqrt{ab}-ab-ab+b\sqrt{ab}}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\left(b\sqrt{a}+a\sqrt{b}\right)-\left(\sqrt{a}^3+\sqrt{b}^3\right)}{a\sqrt{ab}-2ab+b\sqrt{ab}}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\sqrt{ab}\left(\sqrt{b}+\sqrt{a}\right)-\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{ab}\left(a-2\sqrt{ab}+b\right)}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{ab}-a+\sqrt{ab}-b\right)}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)^2}-\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=-\frac{\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)^2}-\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{-\sqrt{a}-\sqrt{b}-ab\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
Tớ làm tới đây thui
a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\) b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)
c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)
d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)
c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)
B=\(\sqrt{3+\sqrt{5}}\)-\(\sqrt{3-\sqrt{5}}\)-\(\sqrt{2}\)
B=\(\sqrt{\frac{1}{2}\left(6+2\sqrt{5}\right)}\)-\(\sqrt{\frac{1}{2}\left(6-2\sqrt{5}\right)}\)-\(\sqrt{2}\)
B=\(\sqrt{\frac{1}{2}\left(5+2\sqrt{5}.1+1\right)}\)-\(\sqrt{\frac{1}{2}\left(5-2\sqrt{5}.1+1\right)}\)-\(\sqrt{2}\)
B=\(\sqrt{\frac{1}{2}\left(\sqrt{5}+1\right)^2}\)-\(\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)^2}\)-\(\sqrt{2}\)
B=\(\frac{\sqrt{5}+1}{\sqrt{2}}\)-\(\frac{\sqrt{5}-1}{\sqrt{2}}\)-\(\sqrt{2}\)
B=\(\frac{2}{\sqrt{2}}\)-\(\sqrt{2}\)
B=\(\sqrt{2}\)-\(\sqrt{2}\)=0
Ta có :
\(B.\sqrt{2}=\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\right).\sqrt{2}\)
\(=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)
\(=\sqrt{5}+1-\left|\sqrt{5}-1\right|-2=0\)
\(\Rightarrow B=0\)