Em tham khảo tại đây nhé:

Câu hỏi của kacura - Toán lớp 8 - Học toán với OnlineMath

2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)

c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)

Câu 1:

\(a^3+a^2b-ab^2-b^3\)

\(=a^2\left(a+b\right)-b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-b^2\right)\)

\(=\left(a+b\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\left(a-b\right)\)

Câu 2:

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+bc^3-a^3b+a^3c-b^3c\)

\(=a\left(b-c\right)\left(b^2+bc+c^2\right)-a^3\left(b-c\right)-bc\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(ab^2+abc+c^2a-a^3-b^2c-bc^2\right)\)

\(=\left(b-c\right)\left[a\left(c-a\right)\left(c+a\right)-b^2\left(c-a\right)-bc\left(c-a\right)\right]\)

\(=\left(b-c\right)\left(c-a\right)\left(ca+a^2-b^2-bc\right)\)

\(=\left(b-c\right)\left(c-a\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

\(\left(a+b+c\right)^2=0\)

\(\Leftrightarrow2ab+2bc+2ac=-2009\)

\(\Leftrightarrow ab+bc+ac=-\dfrac{2009}{2}\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{4036081}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{4036081}{4}\)

\(a^2+b^2+c^2=2009\)

nên \(a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=4036081\)

\(\Leftrightarrow a^4+b^4+c^4=\dfrac{4036081}{2}\)