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#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(2x^3-x^2+5x+3=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
\(=2x^2-6x+x-3=\left(x-3\right)\left(2x+1\right)\)
\(=2x^2-6x+x-3=2x\left(x-3\right)+\left(x-3\right)=\left(2x+1\right)\left(x-3\right)\)
= 2x^2 - x + 6x - 3
= x(2x - 1) + 3(2x - 1)
= (x + 3)(2x - 1)
\(2x^2+5x-3\)
\(=2x^2+6x-x-3\)
\(=2x\left(x+3\right)-\left(x+3\right)\)
\(=\left(2x-1\right)\left(x+3\right)\)
a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)
b) \(=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)
\(-2x^2+5x+3\)
\(=-2x^2+6x-x+3\)
\(=-2x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(-2x-1\right)\)
Lời giải:
$2x^2-5x-6=2(x^2-\frac{5}{2}x+\frac{5^2}{4^2})-\frac{73}{8}$
$=2(x-\frac{5}{4})^2-\frac{73}{8}$
$=2(x-\frac{5}{4}-\frac{\sqrt{73}}{4})(x-\frac{5}{4}+\frac{\sqrt{73}}{4})$