a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;

b ) -(y-x-3)*(y+x+3)

a) \(12x^3-9x^2+3x\)

\(=3x\left(4x^2-3x+1\right)\)

b) \(x^2-y^2+6x+9\)

\(=\left(x^2+6x+9\right)-y^2\)

\(=\left(x+3\right)^2-y^2\)

\(=\left(x+y+3\right)\left(x-y+3\right)\)

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

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

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

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

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

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

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

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

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

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\\ =x\left(x-4\right)\left(x-11\right)\)

\(x^2-16x-15x\left(x-4\right)\)

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2+4x-15x\right)\)

\(=\left(x-4\right)\left(x^2-11x\right)=x\left(x-4\right)\left(x-11\right)\)

\(x^3-16x-15x\left(x-4\right)\)

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2+4x-15x\right)\)

\(=x\left(x-4\right)\left(x-11\right)\)

Đề sai nhé .Sửu lại

\(x^2-4x^2y^2+4+4x\)

\(=\left(x^2+4x+4\right)-4x^2y^2\)

\(=\left(x+2\right)^2-\left(2xy\right)^2\)

\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)

a) \(=\left(a+2c\right)^2-16=\left(a+2c-4\right)\left(a+2c+4\right)\)

b) \(=3y\left(4-x^2\right)+9\left(4-x^2\right)=3\left(4-x^2\right)\left(y+3\right)\)

\(=3\left(2-x\right)\left(2+x\right)\left(y+3\right)\)

a, a² + 4ac + 4c² - 16 = (a + 2c)² - 4² = (a + 2c -4).(a + 2c +4)

b, 12y - 9x² + 36 - 3x²y = (12y + 36) - (3x²y + 9x²) = 12.(y+ 3) - 3x².(y + 3) =(y + 3).(12 - 3x²)

=8x^2+12x=4x(2x+3)

\(x^2+7x^2+12x=8x^2+12x=4x\left(2x+3\right)\)

\(12x^2+7x-12=12x^2-5x+12x-12\)

\(=x\left(12x-5\right)+12\left(x-1\right)\)

Đề sai rồi bạn ời

\(12x^2+7x-12=\left(12x^2-9x\right)+\left(16x-12\right)=3x\left(4x-3\right)+4\left(4x-3\right)=\left(3x+4\right)\left(4x-3\right)\)

Phân tích đa thức sau thành nhân tử 7x^2 +12x+5