Thừa số tổng quát:
\(1+\dfrac{1}{n^2+2n}=\dfrac{n^2+2n+1}{n^2+2n}=\dfrac{\left(n+1\right)^2}{\left(n+1\right)^2-1}\)
Đặt: \(\left(n+1\right)^2=t\ge0\) biểu thức được phát biểu dưới dạng: \(\dfrac{t}{t-1}\) Thay vào bài toán tìm được giá trị.
Thừa số tổng quát:
\(1+\dfrac{1}{n^2+2n}=\dfrac{n^2+2n+1}{n^2+2n}=\dfrac{\left(n+1\right)^2}{\left(n+1\right)^2-1}\)
Đặt: \(\left(n+1\right)^2=t\ge0\) biểu thức được phát biểu dưới dạng: \(\dfrac{t}{t-1}\) Thay vào bài toán tìm được giá trị.
Rút gọn bt: A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+\dfrac{n-3}{3}+..+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Tính các tổng sau :
a, \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
b, \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.11}+........+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)
c,\(\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.22}+....+\dfrac{1}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)
chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
Rút gọn
A=\(\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)...\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right)...\left(12^4+\dfrac{1}{4}\right)}\)
Làm tính nhanh (làm kĩ giúp ạ)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}\)
Giải phương trình:
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2\)
Tính giá trị biểu thức :
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right).....\left(29^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right).....\left(30^4+\dfrac{1}{4}\right)}\)
T^T
giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)