Chứng minh \(E=\left(\sqrt[3]{2}+1\right)\cdot\sqrt[3]{\dfrac{\sqrt[3]{2}-1}{3}\in Z}\)
1.Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
2.Chứng minh: A= \(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
Chứng minh: A=\(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
Đặt B là tên biểu thức
Với mọi n thuộc N*, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) (*)
Áp dụng (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
1) Chứng minh đẳng thức $\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right) \cdot \sqrt{3+2 \sqrt{2}}=-4$.
2) Rút gọn biểu thức $A=\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right): \dfrac{2}{x+\sqrt{x}-2}$ với $x>0 ; x \neq 1$.
1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)
=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)
=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)
=\(\dfrac{-4\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
=-4
2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)
=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)
1. (1−5+√2√2+1)⋅√3+2√2=−4√2+1√(√2+1)2=−4(1−5+22+1)⋅3+22=−42+1(2+1)2=−4.
2. Với x>0;x≠1x>0;x≠1 ta có:
A=(√xx+√x−1√x−1):2x+√x−2A=(xx+x−1x−1):2x+x−2
⇔A=(√x√x(√x+1)−1√x−1):2(√x−1)(√x+2)⇔A=(xx(x+1)−1x−1):2(x−1)(x+2)
⇔A=−2(√x−1)(√x+1)⋅(√x−1)(√x+2)2⇔A=−2(x−1)(x+1)⋅(x−1)(x+2)2
⇔A=−(√x+2)√x+1⇔A=−(x+2)x+1. Vạyy với x>0;x≠1x>0;x≠1, ta có A=−(√x+2)√x+1A=−(x+2)x+1.
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
chứng minh rằng:\(\dfrac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\dfrac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\dfrac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \dfrac{3}{7}\)
Bạn tham khảo câu số 9:
Tính:
\(A=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}-\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(B=\left(1+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\right)\cdot\left(1-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
\(C=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{2}-\sqrt{3}}\)
\(D=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{6}}\)
\(E=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
a: \(A=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)
b: \(B=\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)=1-2=-1\)
c: \(B=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{6}+2\)
Chứng minh rằng:
\(\dfrac{1}{3\left(\sqrt{2}+1\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{1}{7\left(\sqrt{4}+\sqrt{3}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)
\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
\(choP=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)....a,tìm.x\in Z.để.P\in Z...b,tìm.x\in z.để.P\in z...c,tìm.x.để.\left|P\right|=P...d,tìm.x.để.\sqrt{P}>P\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0.5\cdot\left(-2\dfrac{3}{5}\right)\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)