5x3-12x2-14x-4
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Bài 1:
a: \(=\dfrac{2x^4-8x^3+2x^2+2x^3-8x^2+2x+18x^2-72x+18+56x-15}{x^2-4x+1}\)
\(=2x^2+2x+18+\dfrac{56x-15}{x^2-4x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow x^2-16-x^2-2x=0\\ \Leftrightarrow2x=-16\Leftrightarrow x=-8\\ b,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow6x\left(1-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ d,\Leftrightarrow12x-56x^2+x^2-16=0\\ \Leftrightarrow55x^2-12x+16=0\\ \Delta=144-4\cdot55\cdot16< 0\\ \Leftrightarrow x\in\varnothing\)
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\(a,=5\left(x^3+8y\right)\\ b,=\left(4x+y\right)^2-16=\left(4x+y-4\right)\left(4x+y+4\right)\\ c,=3\left(x^2+2\cdot\dfrac{7}{3}x-5\right)\\ =3\left(x^2+2\cdot\dfrac{7}{3}x+\dfrac{49}{9}-\dfrac{94}{9}\right)\\ =3\left(x+\dfrac{7}{3}-\dfrac{\sqrt{94}}{3}\right)\left(x+\dfrac{7}{3}+\dfrac{\sqrt{94}}{3}\right)\)
a: \(5x^3+40y=5\left(x^3+8y\right)\)
b: \(16x^2+8xy+y^2-16\)
\(=\left(4x+y\right)^2-16\)
\(=\left(4x+y-4\right)\left(4x+y+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(25x^2y-10xy+30xy^2=xy\left(25x+30y-10\right)\)
b) \(16x^2-20x=4x\left(4x-5\right)\)
c) \(18x^3+12x^2-30x=x\left(18x^2+12x-30\right)=x\left(x-1\right)\left(18x+9\right)=96x\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)
d) \(14x\left(x+y\right)-4y\left(x+y\right)=\left(14x-4y\right)\left(x+y\right)\)
e) \(9x^3y^2+12xy^2+15x^2y^2=3xy^2\left(3x^2+5x+4\right)\)
f) \(x\left(x-3\right)-y\left(3-x\right)=x\left(x-3\right)+y\left(x-3\right)=\left(x+y\right)\left(x-3\right)\)
g) \(5x\left(x+4\right)-2y\left(4+x\right)=\left(x+4\right)\left(5x-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
![](https://rs.olm.vn/images/avt/0.png?1311)
f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+5=t\), ta có :
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
Thay và ta có :
\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)