Violympic toán 9
Các câu hỏi tương tự
cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}..............\frac{2n-1}{2n}\)
Chứng minh A<\(\frac{1}{\sqrt{3n+1}}\)
1.Rút gọn
\(A=\left(\frac{2\sqrt[3]{2xy}}{x^2y^2-\sqrt[3]{4}}+\frac{xy-\sqrt[3]{2}}{2xy+2\sqrt[3]{2}}\right)\cdot\frac{2xy}{xy+\sqrt[3]{2}}-\frac{xy}{xy-\sqrt[3]{2}}\)
2. Chứng minh
\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+\left(2n-1\right)^4}=\frac{n^2}{4n^2+1}\)
tính S=\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{1+3}}+\frac{1}{\sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5}}+...+\frac{1}{\sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5+...+\left(2n+1\right)}}\)
1,Rút gọn:
a, \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+2}\)
b,\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
tìm x, biết :
a, \(1-(5\frac{3}{8}+x-7\frac{5}{24}):(-16\frac{2}{3})=0\)
b, \((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
c, \(1\frac{3}{5}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}.x=\frac{16}{5}\)
Cho a,b,c>=1.
CMR: \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}>=\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
cho a, b, c ≥ 1
cmr: \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
a, \(\frac{3}{5}.x-\frac{1}{2}=\frac{1}{7}\)
b, \(\frac{1}{4}+\frac{1}{3}:3x=-5\)
c, \(\frac{1}{3}.x+\frac{2}{5}\left(x+1\right)=0\)
d, \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
Chmr ta luôn có:
\(P_n=\frac{1.3.5...\left(2n-1\right)}{2.4.6...2n}< \frac{1}{\sqrt{2n+1}};\forall n\in Z\)