ok ; f) ... = \(\left(x^2+y^2-17\right)^2-\left(2xy-8\right)^2\)

\(=\left(x^2+y^2-17-2xy+8\right)\left(x^2+y^2-17+2xy-8\right)\)

\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-25\right]\)

\(=\left(x-y+3\right)\left(x-y-3\right)\left(x+y-5\right)\left(x+y+5\right)\)

ĐKXĐ : \(x\ne-4;-5;-6;-7\)

P/t \(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)  \(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)  \(\Leftrightarrow x^2+11x-26=0\)  

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

ĐKXĐ : \(x>0\)

\(0=0.1;2=1.2;6=2.3;12=3.4;...\) ; \(9900=99.100\)

Tập hợp G có số phần tử : \(\left(99-0\right):1+1=100\) (p/t) 

x>0 ; x<0 

\(S=...=\dfrac{\left(99-1\right):2+1}{2}.\left(99+1\right)=2500\)

b) ... = \(\left(7x-4-2x+1\right)\left(7x-4+2x-1\right)\)  = \(\left(5x-3\right)\left(9x-5\right)\)

d) ... = \(\left(x^2+6x+9\right)-\left(y^2-2y+1\right)=\left(x+3\right)^2-\left(y-1\right)^2=\left(x+3-y+1\right)\left(x+3+y-1\right)\)

\(=\left(x-y+4\right)\left(x+y+2\right)\)

Độ dài đường chéo thứ 2 là : \(\dfrac{40.4}{5}=32\left(m\right)\)

Diện tích mảnh đất : \(\dfrac{1}{2}.40.32=640\left(m^2\right)\)

Trên mảnh đất thu hoạch được số rau : \(640:10.20=1280\) ( yến )  = 128 tạ 

\(ĐKXĐ:1\ne x\ge0\)

... = \(\left[\dfrac{\sqrt{x}+3}{1-x}-\dfrac{3\left(1+\sqrt{x}\right)}{1-x}\right]:\dfrac{x-2}{x-1}\)  = \(\dfrac{-2\sqrt{x}}{1-x}.\dfrac{x-1}{x-2}=\dfrac{2\sqrt{x}}{x-2}\)

\(Q=\dfrac{3}{\left(x+\sqrt{2}\right)^2+5}\le\dfrac{3}{5}.'='\Leftrightarrow x=-\sqrt{2}\)