\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}.\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}< 1\)

ds :2108408.33333333

\(sin^2a+cos^2a-sin^4a-2cos^2a+sin^2a\)

\(=2sin^2a-cos^2a-sin^4a\)

\(=2sin^2a-cos^2a-\left(\frac{1-cos2a}{2}\right)^2\)

khai triển ra rồi quy đồng lên

\(=\frac{8sin^2a-4cos^2a-1+2cos2a-cos^22a}{4}\)

Mà \(2cos2a=2\left(cos^2a-1\right)=4cos^2-2\)

\(\Rightarrow\frac{8sin^2a-cos^22a-3}{4}\)

Mà \(-cos^22a=sin^22a-1=4sin^2cos^2-1\)

\(\Rightarrow\frac{8sin^2a+4sin^2a.cos^2a-4}{4}\)

\(=\frac{4sin^2a.\left(2-cos^2a\right)-4}{4}\)

\(=sin^2a\left(1+sin^2a\right)-1\)

\(=sin^4a-cos^2a\)

viết lại đề đi cậu ơi

a) \(ĐKXĐ:x\ne\pm3\)

b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)

\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)

\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)

\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)

\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)

\(\Leftrightarrow A=\frac{-3}{x+3}\)

c) Tại \(x=-\frac{1}{2}\)

\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)

\(\Leftrightarrow A=\frac{-6}{5}\)

d) Để \(A>0\)

\(\Leftrightarrow\frac{-3}{x+3}>0\)

\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)

\(\Leftrightarrow x< -3\)

e) +) Với \(A>\frac{-1}{2}\)

\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)

\(\Leftrightarrow-6>-x-3\)

\(\Leftrightarrow x>3\)(tm)

+) Với \(A< -\frac{1}{2}\)

\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)

\(\Leftrightarrow-6< -x-3\)

\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))

+) Với \(A=-\frac{1}{2}\)

\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)

\(\Leftrightarrow x+3=6\)

\(\Leftrightarrow x=3\)(ktm)

Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)

Hihi