7: Ta có: \(A=x^2+2xy-3y^2\)
\(=x^2+3xy-xy-3y^2\)
\(=x\left(x+3y\right)-y\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x-y\right)\)
\(1,A=x^4+\dfrac{1}{4}=x^4+x^2+\dfrac{1}{4}-x^2\\ =\left(x^2+\dfrac{1}{4}\right)^2-x^2\\ =\left(x^2+x+\dfrac{1}{4}\right)\left(x^2-x+\dfrac{1}{4}\right)\\ =\left(x+\dfrac{1}{2}\right)^2\left(x-\dfrac{1}{2}\right)^2\\ 2,A=x^4+4=\left(x^2+2\right)^2-4x^2\\ =\left(x^2-2x+2\right)\left(x^2+2x+2\right)\\ 3,A=4x^4+16=4\left(x^4+4x^2+16\right)-16x^2\\ =4\left(x^2+4\right)^2-16x^2\\ =\left(2x^2-16x+8\right)\left(2x^2+16x+8\right)\\ 4,A=4x^4+\dfrac{1}{16}=\left(4x^4+x^2+\dfrac{1}{16}\right)-x^2\\ =\left(2x^2+\dfrac{1}{8}\right)^2-x^2\)
\(=\left(2x^2-x+\dfrac{1}{8}\right)\left(2x^2+x+\dfrac{1}{8}\right)\)
Giúp e bài 2 thôi ạ bài 1 e làm r ạ! Mong mn giúp e, e cần gấp ạ!
