x3+3x2-10x=0

=>x(3+3.2-10)=0

=>x=0

x3-5x2-14x=0

=>x(3-5.2-14)=0

=>x=0

x3+5x2-24x=0

=>x(3+5.2-24)=0

=>x=0

Câu a)

\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)

\(\Rightarrow x=0;x=5;x=2\)

ĐKXĐ: x>=0; y>=1; 2y-3x-4>=0

\(\begin{cases} x^3 + x^2 + y^2 - x^2y - xy - y = 0 \quad (1) \\ \sqrt{x} + \sqrt{y - 1} = \sqrt{2y - 3x - 4} \quad (2) \end{cases}\)

(1): \(x^3 + x^2 + y^2 - x^2y - xy - y = 0\)

=>\((x^3+x^2)-(x^2y+xy)+(y^2-y)=0\)

=>\(x^2(x+1)-xy(x+1)+y(y-1)=0\)

=>\(y^2 - (x^2 + x + 1)y + (x^3 + x^2) = 0\)

\(\Delta = (x^2 + x + 1)^2 - 4 \cdot 1 \cdot (x^3 + x^2)\)

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

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

\(=x^4-2x^3-x^2+2x+1=(x^2-x-1)^2\) >=0∀x

=>(1) có hai nghiệm là:

\(\left[\begin{array}{l}y=\frac{(x^2+x+1)+(x^2-x-1)}{2}=\frac{x^2+x+1+x^2-x-1}{2}=\frac{2x^2}{2}=x^2\\ y=\frac{(x^2+x+1)-(x^2-x-1)}{2}=\frac{x^2+x+1-x^2+x+1}{2}=\frac{2x+2}{2}=x+1\end{array}\right.\)

TH1: \(y = x^2\)

(2) sẽ trở thành: \(\sqrt{x} + \sqrt{x^2 - 1} = \sqrt{2x^2 - 3x - 4}\)

=>\(x + x^2 - 1 + 2\sqrt{x(x^2 - 1)} = 2x^2 - 3x - 4\)

=>\(2\sqrt{x^3 - x}=x^2-4x-3\)

=>\(\begin{cases}4(x^3-x)=(x^2-4x-3)^2\\ x^2-4x-3\ge0\end{cases}\)

=>\(\begin{cases}4x^3-4x=x^4+16x^2+9-8x^3-6x^2+24x\\ x^2-4x+4-7\ge0\end{cases}\)

=>\(\begin{cases}x^4-12x^3+10x^2+28x+9=0\\ \left(x-2\right)^2\ge7\end{cases}\Rightarrow\begin{cases}(x^2-10x-9)(x^2-2x-1)=0\\ \left(x-2\right)^2\ge7\end{cases}\)

=>\(x=5+\sqrt{34}\)

=>\(y=x^2=(5+\sqrt{34})^2=59+10\sqrt{34}\)

TH2: y=x+1

Thay y=x+1 vào (2), ta được:

\(\sqrt{x} + \sqrt{(x + 1) - 1} = \sqrt{2(x + 1) - 3x - 4}\)

=>\(\sqrt{x}+\sqrt{x}=\sqrt{2x + 2 - 3x - 4}\)

=>\(2\sqrt{x}=\sqrt{-x-2}\)

=>-x-2>=0 và 4x=-x-2

=>-x>=2 và 5x=-2

=>x<=-2 và x=-2/5

=>x∈∅

=>Loại

6: \(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\)

\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}\)

\(=\frac{\left(2x-5\right)\left(x^2+3\right)}{2x-5}=x^2+3\)

2: \(\frac{2x^4-5x^2+x^3-3-3x}{x^2-3}\)

\(=\frac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)

\(=\frac{2x^2\left(x^2-3\right)+x\cdot\left(x^2-3\right)+\left(x^2-3\right)}{x^2-3}=2x^2+x+1\)

5: \(\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)

\(=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}=\frac{\left(2x^2-x+1\right)\left(x+3\right)}{2x^2-x+1}\)

=x+3

3: \(\left(x-y-z\right)^5:\left(x-y-z\right)^3=\left(x-y-z\right)^{5-3}=\left(x-y-z\right)^2\)

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

\(=\frac{x^2\left(x-3\right)+\left(x-3\right)}{x-3}=x^2+1\)

a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)

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

\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)

\(=-\left(2x-4\right)\left(x+8\right)\)

b) \(x^3+x^2y-15x-15y\)

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

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

c) \(3\left(x+8\right)-x^2-8x\)

\(=3\left(x+8\right)-x\left(x+8\right)\)

\(=\left(x+8\right)\left(3-x\right)\)

d) \(x^3-3x^2+1-3x\)

\(=x^3+1-3x^2-3x\)

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

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

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

d) \(5x^2-5y^2-20x+20y\)

\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y-4\right)\)

a.

⇔ \(5x^2-3x+\left(-7\right)-1=0\)

⇔ \(5x^2-3x-8=0\)

Δ=\(b^2-4ac\) \(=\left(-3\right)^2-4.5.\left(-8\right)=169\)>0

Vì Δ>0 nên pt có 2 nghiệm phân biệt:

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{169}}{2.5}=\dfrac{8}{5}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{169}}{2.5}=-1\)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

x²-x-5x+5=0

<=>x(x-1)-5(x-1)=0

<=>(x-5)(x-1)=0

<=>x=5 hoặc x=1

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