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a) TA CÓ:
\(a^2bc^2d-ab^2cd^2+a^2bcd^2-ab^2c^2d\)
\(=abcd\left(ac-bd+ad-bc\right)\)
\(=abcd\left[a\left(c+d\right)-b\left(c+d\right)\right]\)
\(=abcd\left(c+d\right)\left(a-b\right)\)
\(x\left(y^2-z^2\right)+z\left(x^2-y^2\right)+y\left(z^2-x^2\right)\)
\(=x\left(y^2-z^2\right)-\left(y^2-z^2+z^2-x^2\right)z+y\left(z^2-x^2\right)\)
\(=x\left(y^2-z^2\right)-z\left(y^2-z^2\right)-z\left(z^2-x^2\right)+y\left(z^2-x^2\right)\)
\(=\left(y^2-z^2\right)\left(x-z\right)+\left(z^2-x^2\right)\left(y-z\right)\)
\(=\left(y-z\right)\left(z-x\right)\left(-\left(y+z\right)+z+x\right)\)
= \(\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
\(A=a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]+4abc\)
\(=a\left(b-c+a\right)\left(b-c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)+4abc\)
\(=\left(a+b-c\right)\left(ab-ac-a^2-bc+ab-b^2\right)+c\left(a^2-2ab+b^2-c^2+4ab\right)\)
\(=\left(a+b-c\right)\left[-c\left(a+b\right)-\left(a-b\right)^2\right]+c\left[\left(a+b\right)^2-c^2\right]\)
\(=\left(a+b-c\right)\left(-ca-cb-a^2+2ab-b^2+ac+cb+c^2\right)\)
\(=\left(a+b-c\right)\left(c^2-\left(a-b\right)^2\right)\)
\(=\left(a+b-c\right)\left(c+a-b\right)\left(a+b-c\right)\)
\(a,a^3-7a-6\)
\(\Leftrightarrow a^3+a^2-a^2-a-6a-6\)
\(\Leftrightarrow a^2\left(a+1\right)-a\left(a+1\right)-6\left(a+1\right)\)
\(\Leftrightarrow\left(a+1\right)\left(a^2-a-6\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
\(b,a^3+4a^2-7a-10\)
\(\Leftrightarrow a^3+5a^2-a^2-5a-2a-10\)
\(\Leftrightarrow a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)
\(\Leftrightarrow\left(a+5\right)\left(a+1\right)\left(a-2\right)\)
\(d,\left(a^2+a\right)^2+4\left(a^2+a\right)-12\)
Đặt a^2+a=y ta có
y^2+4y-12=(y+6)(y-2)
<=> (a^2+a+6)(a^2+a-2)
<=> (a^2+a+6)(x-1)(x+2)
Chtt nha chẳng em mới học lớp 6