Violympic toán 9
Các câu hỏi tương tự
Tính các tổng sau:
Adfrac{1}{2.5}+dfrac{1}{5.8}+dfrac{1}{8.11}+.....+dfrac{1}{left(3n-1right)left(3n+2right)}
Bdfrac{1}{1.3.5}+dfrac{1}{3.5.7}+dfrac{1}{5.7.9}+....+dfrac{1}{left(2n-1right)left(2n+1right)left(2n+3right)}
Csqrt{1+dfrac{1}{1^2}+dfrac{1}{2^2}}+sqrt{1+dfrac{1}{2^2}+dfrac{1}{3^2}}+sqrt{1+dfrac{1}{3^2}+dfrac{1}{4^2}}+....+sqrt{1+dfrac{1}{2018^2}+dfrac{1}{2019^2}}
Đọc tiếp
Tính các tổng sau:
A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
B=\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}\)
C=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+....+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
cho hai biểu thức A=\(\dfrac{2\sqrt{x}-4}{\sqrt{x}-1}\) và B=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) với x\(\ge\)0, x\(\ne\)1
a.tính giá trị của A khi x=4
b.rút gọn B
c.so sánh A.B với 5
Chứng minh: Adfrac{2^3+1}{2^3-1}.dfrac{3^3+1}{3^3-1}.dfrac{4^3+1}{4^3-1}....dfrac{9^3+1}{9^3-1} dfrac{3}{2}
Bdfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+....+dfrac{1}{n!} 1
Cdfrac{1}{2!}+dfrac{2}{3!}+dfrac{3}{4!}+....+dfrac{n-1}{n!} 1
Dleft(1-dfrac{2}{6}right)left(1-dfrac{2}{12}right)left(1-dfrac{2}{20}right)....left(1-dfrac{2}{nleft(n+1right)}right)dfrac{1}{3}
Đọc tiếp
Chứng minh: \(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}.\dfrac{4^3+1}{4^3-1}....\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}\)
\(B=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{n!}< 1\)
\(C=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+....+\dfrac{n-1}{n!}< 1\)
D=\(\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)....\left(1-\dfrac{2}{n\left(n+1\right)}\right)>\dfrac{1}{3}\)
So sánh A với 1
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
a)tính tổng S=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+..+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\)
b)Áp dụng, tìm phần nguyên của A=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\) với n lẻ
CM: \(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}+\dfrac{3}{\sqrt{6}-3}\right).\dfrac{5}{9\sqrt{6}+4}=\dfrac{1}{2}\)
Cho \(a,b,c>0\) thỏa mãn \(a^4+b^4+c^4=3\). Chứng minh:
\(\dfrac{a^2}{b^3+1}+\dfrac{b^2}{c^3+1}+\dfrac{c^2}{a^3+1}\ge\dfrac{3}{2}\)
a,Rút gọn :
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)...\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right)...\left(52^4+\dfrac{1}{4}\right)}\)
b, Tìm nghiệm nguyên: \(4x^2-8y^3+2z^2+4x-4=0\)