\(a,x^2+27=\left(\sqrt[3]{x^2}\right)^3+3^3=\left(\sqrt[3]{x^2}+3\right)\left[\left(\sqrt[3]{x^2}\right)^2-3\sqrt[3]{x^2}+9\right]\)
\(g,x^2+2x-3=\left(x^2+2x+1\right)-4=\left(x+1\right)^2-2^2=\left(x+1+2\right)\left(x+1-2\right)=\left(x+3\right)\left(x-1\right)\)
\(i,4x^2y^2-\left(x^2+y^2\right)^2=\left(2xy\right)^2-\left(x^2+y^2\right)^2=\left(2xy+x^2+y^2\right)\left(2xy-x^2+y^2\right)=\left(x+y\right)^2\left(2xy-x^2+y^2\right)\)
a) \(x^3+27\) = \(x^3+3^3=\left(x+3\right)\left(x^2-3x+3^2\right)\)
= \(\left(x+3\right)\left(x^2-3x+9\right)\)
câu a nếu của lp 8 là như thế này
câu b hình như đề sai
c) \(x^2+2x-3\)
= \(x^2-x+3x-3\)
= \(x\left(x-1\right)+3\left(x-1\right)\)
= \(\left(x+3\right)\left(x-1\right)\)
i) \(4x^2y^2-\left(x^2+y^2\right)^2\)
= \(\left(2xy\right)^2-\left(x^2+y^2\right)^2\)
= \(\left(2xy+x^2+y^2\right)\left(2xy-x^2-y^2\right)\)
để chút mk lm tiếp busy rồi
\(b,a^2-b^2-4a+4=\left(a^2-4a+4\right)-b^2=\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)