f) \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow x^3-5x^2+6x-x^2+5x-6=0\)

\(\Leftrightarrow x\left(x^2-5x+6\right)-\left(x-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-3x+6\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\)x = 1 hoặc x = 2 hoặc x = 3

g) +) Với x\(\ge\)0,5 thì |2x - 1| = 2x - 1

Phương trình trở thành: x + 2x - 1 =5

<=> 3x - 1 = 5  

<=> x = 2 > 0,5 (thỏa mãn)

+) Với x < 0,5 thì |2x - 1| = 1 - 2x

Phương trình trở thành: x + 1 - 2x = 5

<=> -x + 1 = 5

<=> x = -4 < 0,5(thỏa mãn)

h) \(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow2\left(x^3+\frac{3}{2}x^2-16x-24\right)=0\)

\(\Leftrightarrow2\left[x^2\left(x+\frac{3}{2}\right)-16\left(x+\frac{3}{2}\right)\right]=0\)

\(\Leftrightarrow2\left(x^2-16\right)\left(x+\frac{3}{2}\right)=0\)

\(\Leftrightarrow2\left(x-4\right)\left(x+4\right)\left(x+\frac{3}{2}\right)=0\)

<=> x = 4 hoặc x = -4 hoặc x = \(\frac{-3}{2}\) 

a, 2x+x=3+5+4

3x=12

x=4

b, 2-2x=2018

-2x=2016

x=-1003

\(3-x+5=2x-4\)

\(\Rightarrow-x-2x=-4-5-3\)

\(\Rightarrow-3x=-12\)

\(\Rightarrow x=4\)

\(2\left|1-x\right|=2018\)

\(\Rightarrow\left|1-x\right|=1009\)

\(\Leftrightarrow\orbr{\begin{cases}1-x=1009\\1-x=-1009\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1008\\x=1010\end{cases}}\)

a, 2x+x=3+5+4

3x=12

x=4

b, 2-2x=2018

-2x=2016

x=-1003

\(x\ge-\frac{1}{2}\Rightarrow3x-2x-1=0\Rightarrow x=1\)

\(x< \frac{-1}{2}\Rightarrow3x+2x+1\Rightarrow x=-\frac{1}{5}\left(loai\right)\)

\(3x-|2x-1|=2\Leftrightarrow|2x-1|=2-3x\)

\(\Rightarrow-2x+1=2-3x\)hoặc \(-2x+1=3x-2\)

\(\Rightarrow1x+1=2\)hoặc \(-5x+1=-2\)

\(\Rightarrow x=1\)hoặc\(x=\frac{5}{3}\)

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

b) 2x(x + 5) - 3(5 + x) = 0

(x + 5)(2x - 3) = 0

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

(x - 1)(x - 2)(x - 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3

a) Ta có: \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy: S={2;-3}

a) Rút gọn E Þ đpcm.

b) Điều kiện xác định E là: x ≠    ± 1  

Rút gọn F ta thu được F = 4 Þ đpcm