xem lại đề: không thấy quy luật của tử:
nội suy: Tử là tích=[(2n+2)(2n+5)+2] với n={1...49}
Tuy nhiên mình không thích sửa đề, đề thế nào làm vậy.%
\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}=...\)
\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}\)
Mấy bạn giúp mình được không :((
Tính nhanh giá trị của biểu thức:
\(A=\dfrac{\left(2^4+2^2+1\right)\left(4^4+4^2+1\right)\left(6^4+6^2+1\right)\left(8^4+8^2+1\right)\left(10^4+10^2+1\right)}{\left(3^4+3^2+1\right)\left(5^4+5^2+1\right)\left(7^4+7^2+1\right)\left(9^4+9^2+1\right)\left(11^4+11^2+1\right)}\)
Giaỉ PT
a) \(\left(\frac{x-1}{99}+x-99\right)+\left(\frac{x-3}{97}+\frac{x-7}{93}\right)+\left(\frac{x-5}{95}+\frac{x-95}{5}\right)=6\)
b) \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
c) \(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}-\frac{2}{x+3}\)
Help !!
Giải pt: \(\left(\frac{x+3}{x-2}\right)^2+6\left(\frac{x-3}{x+2}\right)^2=\frac{7\left(x^2-9\right)}{x^2-4}\)
A=\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\)
giải pt
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải PT
\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\)
