Tính GTBT :A = \(\sqrt{1+\frac{2}{3}}\cdot\sqrt{1+\frac{2}{4}}\cdot...\cdot\sqrt{1+\frac{1}{2005}}\cdot\sqrt{1+\frac{2}{2006}}\)
\(y=\frac{1}{2+\sqrt{2}}+\frac{1}{3\cdot\sqrt{2}+2\cdot\sqrt{3}}+\frac{1}{4\cdot\sqrt{3}+3\cdot\sqrt{4}}+...+\frac{1}{100\cdot\sqrt{99}+99\cdot\sqrt{100}}\)tính y
\(y=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(y=1-\frac{1}{10}=\frac{9}{10}\)
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:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
1. Tính:
a. \(\text{[}\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\sqrt{\frac{1}{ab}}\text{]}\cdot\sqrt{ab}\)
b.\(\text{[}-\frac{am}{b}\cdot\sqrt{\frac{n}{m}}-\frac{ab}{n}\cdot\sqrt{mn}+\frac{a^2}{b^2}\cdot\sqrt{\frac{m}{n}}\text{]}\cdot\text{[}a^2b^2\cdot\sqrt{\frac{n}{m}}\text{]}\)
a) \(\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}+\frac{1}{\sqrt{ab}}\right).\sqrt{ab}\) (ĐK : \(\hept{\begin{cases}a>0\\b>0\end{cases}}\)hoặc \(\hept{\begin{cases}a< 0\\b< 0\end{cases}}\))
\(=ab+2b-a+1\)
b) \(\left(-\frac{am}{b}\sqrt{\frac{n}{m}}-\frac{ab}{n}.\sqrt{mn}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\) (ĐK bạn tự xét nhé ^^)
\(=\left(-\frac{a\sqrt{mn}}{b}-\frac{ab\sqrt{m}}{\sqrt{n}}+\frac{a^2}{b^2}.\sqrt{\frac{m}{n}}\right)\left(a^2b^2.\sqrt{\frac{n}{m}}\right)\)
\(=a^2b^2\left(\frac{-an}{b}-ab+\frac{a^2}{b^2}\right)=-a^3bn-a^3b^3+a^4=a^3\left(a-bn-b^3\right)\)
tính: \(\text{[}\sqrt{2}-1\text{]}^2-\frac{3}{2}\cdot\sqrt{\text{[}-2\text{]}^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}\cdot\sqrt{2}\)
Chứng minh: \(\sqrt{x}\cdot\text{[}1-\sqrt{x}\text{]}\le\frac{1}{4}v\text{ới}x\ge0\)
TÍNH : \(\left(\sqrt{2}-1\right)^2-\frac{3}{2}\sqrt{\left(-2\right)^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.\sqrt{2}\)
\(=\left(\sqrt{2}-1\right)^2-\frac{3}{2}.2+\frac{4\sqrt{2}}{5}+\sqrt{\frac{36}{25}}.\sqrt{2}\)
\(=3-2\sqrt{2}-3+\frac{4\sqrt{2}}{5}+\frac{6\sqrt{2}}{5}=\frac{10\sqrt{2}}{5}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0\)
CHỨNG MINH :
Ta có : \(\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right]+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)với mọi \(x\ge0\)
Vậy ta có điều phải chứng minh.
\(A=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{2}\right)\cdot\frac{5}{7}}\)
1. CHỨNG MINH ĐẲNG THỨC
a. \(\text{[}3+2\sqrt{6}-\sqrt{33}\text{]}\cdot\text{[}\sqrt{22}+\sqrt{6}+4\text{]}=24\)
b. \(\text{[}\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\text{]}\cdot\text{[}15+2\sqrt{6}\text{]}\)
c.\(\text{[}\frac{4}{3}\cdot\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\text{]}\cdot\text{[}\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\text{]}=4\)
d. \(\sqrt{\text{[}1-\sqrt{1989}\text{]}^2}\cdot\sqrt{1990+2\sqrt{1989}}=1988\)
e. \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)với \(a>0;b>0\)và \(a\ne b\)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right)\cdot\frac{5}{7}}\)