a) x³ - 7x + 6 = 0
x³ - x - 6x + 6 = 0
(x³ - x) - (6x - 6) = 0
x(x² - 1) - 6(x - 1) = 0
x(x - 1)(x + 1) - 6(x - 1) = 0
(x - 1)[x(x + 1) - 6] = 0
(x - 1)(x² + x - 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 hoăkc x + 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x + 3 = 0
x = -3
Vậy x = -3; x = 1; x = 2
a: \(x^3-7x+6=0\)
=>\(x^3-x-6x+6=0\)
=>\(x\left(x^2-1\right)-6\left(x-1\right)=0\)
=>x(x-1)(x+1)-6(x-1)=0
=>(x-1)(x^2+x-6)=0
=>(x-1)(x+3)(x-2)=0
=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)
b: \(x^4+4x^2-5=0\)
=>\(x^4+5x^2-x^2-5=0\)
=>\(\left(x^2+5\right)\left(x^2-1\right)=0\)
=>\(x^2-1=0\)
=>\(x^2=1\)
=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)
c: \(x^4+x^3-x^2-x=0\)
=>\(x^3\left(x+1\right)-x\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^3-x\right)=0\)
=>\(x\left(x+1\right)^2\cdot\left(x-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=1\end{array}\right.\)
d: \(x^2+6x-x-6=0\)
=>x(x+6)-(x+6)=0
=>(x+6)(x-1)=0
=>\(\left[\begin{array}{l}x+6=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=1\end{array}\right.\)
e: \(x^2-4x+5x-20=0\)
=>x(x-4)+5(x-4)=0
=>(x-4)(x+5)=0
=>\(\left[\begin{array}{l}x-4=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-5\end{array}\right.\)
f: \(x^2-10x+2x-20=0\)
=>x(x-10)+2(x-10)=0
=>(x-10)(x+2)=0
=>\(\left[\begin{array}{l}x-10=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-2\end{array}\right.\)
g: \(x^4-x^3-x^2+1=0\)
=>\(x^3\left(x-1\right)-\left(x^2-1\right)=0\)
=>\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x-1\right)=0\)
TH1: x-1=0
=>x=1
TH2: \(x^3-x-1=0\)
=>x≃1,32
h: \(x^5+x^4+x^3+x^2+x+1=0\)
=>\(x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)=0\)
=>\(\left(x^2+x+1\right)\left(x^3+1\right)=0\)
mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)
nên \(x^3+1=0\)
=>\(x^3=-1\)
=>x=-1
i: \(x^2-9+\left(x+3\right)\left(3x-5\right)=0\)
=>(x-3)(x+3)+(x+3)(3x-5)=0
=>(x+3)(x-3+3x-5)=0
=>(x+3)(4x-8)=0
=>4(x+3)(x-2)=0
=>(x+3)(x-2)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)
j: \(64x^2-9+8x+3=0\)
=>(8x+3)(8x-3)+(8x+3)=0
=>(8x+3)(8x-3+1)=0
=>(8x+3)(8x-2)=0
=>\(\left[\begin{array}{l}8x+3=0\\ 8x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac38\\ x=\frac28=\frac14\end{array}\right.\)
