\(\frac{1}{\sqrt{k\left(2018-k+1\right)}}>\frac{2}{k+2019-k}=\frac{2}{2019}\)
Ap dụng bài toan được
\(A>\frac{2}{2019}+\frac{2}{2019}+...+\frac{2}{2019}=2.\frac{2018}{2019}\)
cho biểu thức A=\(\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}\)
chứng tỏ A>\(\sqrt{\frac{2018}{2019}}\)
Tính:
A) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}-2\right)^2\)
B) \(\left(11-4\sqrt{3}\right)\left(11-4\sqrt{3}\right)\)
C) \(\left(1+\sqrt{2018}\right)\left(\sqrt{2019}-2\sqrt{2018}\right)\)
D)
\(\left(\sqrt{2}-1\right)^2+\frac{3}{2}\sqrt{\left(-2\right)}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.2\)
CMR : \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2018}}>2\left(\sqrt{2018}-1\right)\)
Chứng minh rằng
a) Với mọi số nguyên dương n có \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+..+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
b) \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}< \sqrt{2017}+\sqrt{2018}\)
Hộ mình vs
Rút gọn \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2018}+\sqrt{2019}}{1+\sqrt{2018}+\sqrt{2019}}\)
So sánh \(\sqrt{2019^2-1}-\sqrt{2018^2-1}\) và \(\frac{2.2018}{\sqrt{2019^2-1}+\sqrt{2018^2-1}}\)
Tính:
A= \(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+ \(\frac{1}{3\sqrt{2}+2\sqrt{3}}\)+....+ \(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
