Câu 1:
\(1.\\ a.x^4+2x^2y+y^2-9\\ =\left(x^2+y\right)^2-3^2\\ =\left(x^2+y-3\right)\left(x^2+y+3\right)\\ b.\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-5^2\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
Câu 2:
a: ĐKXĐ: \(x\notin\left\{-1;1\right\}\)
A=\(\left(\dfrac{1-x^3}{1-x}-x\right):\dfrac{1-x^2}{1-x-x^2+x^3}\)
\(=\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\dfrac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\dfrac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)
\(=\left(x^2+1\right)\cdot\dfrac{\left(1-x\right)\left(1-x^2\right)}{1-x^2}=\left(x^2+1\right)\left(1-x\right)\)
b: \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{1}{9}\)
=>\(\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{1}{3}\\x-\dfrac{2}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)
Khi x=1/3 thì \(A=\left[\left(\dfrac{1}{3}\right)^2+1\right]\left[1-\dfrac{1}{3}\right]=\dfrac{2}{3}\cdot\left(\dfrac{1}{9}+1\right)=\dfrac{2}{3}\cdot\dfrac{10}{9}=\dfrac{20}{27}\)
c: A<0
=>\(\left(x^2+1\right)\left(1-x\right)< 0\)
=>1-x<0
=>x>1
