Tính tổng: \(A=\frac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\frac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\frac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
tinhs tổng:
\(A=\frac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\frac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\frac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{2-2\sqrt{x}+2+2\sqrt{x}}{\left(2+2\sqrt{x}\right)\left(2-2\sqrt{x}\right)}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{4}{4-4x}-\frac{x^2+1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{1+x-x^2-1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)=\frac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}.\frac{x+1}{x}=1\)
Nếu bạn bảo kiểm tra thì lời giải đúng rồi nhé!
Tính D= \(\frac{x^4-\left(x-1\right)^2}{\left(x+1\right)^2-x^2}+\frac{\left(x-1\right)^2-x^2}{1-\left(x^2+x\right)^2}+\frac{\left(x^2-x\right)^2-1}{x^4-\left(x+1^2\right)}\)
Tính nhanh tổng sau:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\)
quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6
=1\x-1\x+6
=6\x(x+6)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)
Tính A = \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+5\right)}+\frac{8}{\left(x+5\right)\left(x+6\right)}\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
Giải phương trình \(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)}+\frac{1}{\left(x^2+1\right)}\)
AYUASGSHXHFSGDB HAGGAHAJF
tính nhanh tổng sau
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+2\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}\)
\(+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\left(n\in N\right)\)
Như vậy,
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}=\frac{x+6}{x.\left(x+6\right)}-\frac{x}{x.\left(x+6\right)}=\frac{6}{x^2+6x}\)
Tính các tổng sau:
a,A=\(\frac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\frac{x^2-\left(x^2-1\right)^2}{x^2\cdot\left(x+1\right)^2-1}+\frac{x^2\cdot\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
b,B=\(\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
Giúp mình với!
a) Chứng minh: \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x\left(x+1\right)}\)
b). Tính nhẩm tổng sau: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)