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b: \(=\dfrac{x^4-x^3-2x^3+2x^2+x^2-x}{x-1}=x^3-2x^2+x\)
Câu 1:
\(x^4+5x^3-12x^2+5x+1=x^4+7x^3+x^2-2x^3-14x^2-x+x^2+7x+1\)
\(=\left(x^4+7x^3+x^2\right)-\left(2x^3+14x^2+x\right)+\left(x^2+7x+1\right)\)
\(=x^2\left(x^2+7x+1\right)-2x\left(x^2+7x+1\right)+\left(x^2+7x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+7x+1\right)\)
\(=\left(x-1\right)^2\left(x^2+7x+1\right)\)
Câu 2:
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=x^4-24x^3+203x^2-720x+900-24x^2\)
\(=x^4-24x^3+179x^2-720x+900\)
\(=\left(x^4-7x^3+30x^2\right)-\left(17x^3-119x^2+510x\right)+\left(30x^2-210x+900\right)\)
\(=x^2\left(x^2-7x+30\right)-17x\left(x^2-7x+30\right)+30\left(x^2-7x+30\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
\(=\left(x^2-2x-15x+30\right)\left(x^2-7x+30\right)\)
\(=\left[x\left(x-2\right)-15\left(x-2\right)\right]\left(x^2-7x+30\right)\)
\(=\left(x-15\right)\left(x-2\right)\left(x^2-7x+30\right)\)
Câu 3:
\(2x^3+11x^2+3x-36=\left(2x^3+14x^2+24x\right)-\left(3x^2+21x+36\right)\)
\(=2x\left(x^2+7x+12\right)-3\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+3x+4x+12\right)\)
\(=\left(2x-3\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=\left(2x-3\right)\left(x+3\right)\left(x+4\right)\)
d, \(2x^3-12x^2+24x-16\)
= 2(\(x^3-6x^2+12x-8\))
=2(x-2)\(^3\)
e, \(x^3-10x^2+25x-9xy^2\)
=\(x\left(x-10x+25-9y^2\right)\)
=\(x\left[\left(x-5\right)^2-\left(3y\right)^2\right]\)
=\(x\left[\left(x-5-3y\right)\left(x-5+3y\right)\right]\)
a, \(x^3-8x^2+16x\)
=\(x^3-4x^2-4x^2+16x\)
= (\(x^3-4x^2\))-\(\left(4x^2-16x\right)\)
=\(x^2\left(x-4\right)-4x\left(x-4\right)\)
=\(\left(x^2-4x^2\right)\left(x-4\right)\)
b, \(3x^2-27\)
=3(\(x^2-9\))
=3\(\left(x^2-3^2\right)\)
=3\(\left(x-3\right)\left(x+3\right)\)
c,\(3x^2-5xy+6x-10y\)
=\(\left(3x^2+6x\right)-\left(5xy+10y\right)\)
=3x(x+2)-5y(x+2)
=(x+2)(3x-5y)
Câu a :
\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
Câu b :
\(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
