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(x - 1)2 + (x - 2)2 = 1 (1)
\(\Leftrightarrow\) x2 - 2x + 1 + x2 - 4x + 4 - 1 = 0
\(\Leftrightarrow\) 2x2 - 6x + 4 = 0
\(\Leftrightarrow\) 2(x2 - 3x + 2) = 0
\(\Leftrightarrow\) x2 - 3x + 2 = 0
\(\Leftrightarrow\) x2 - 2x - x + 2 = 0
\(\Leftrightarrow\) x(x - 2) - (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy S = {2; 1}
x4 - 3x3 + 3x2 - 3x + 2 = 0 (2)
\(\Leftrightarrow\) x4 - 3x3 + 3x2 - x - 2x + 2 = 0
\(\Leftrightarrow\) x(x3 - 3x2 + 3x - 1) - 2(x - 2) = 0
\(\Leftrightarrow\) x(x - 1)3 - 2(x - 1) = 0
\(\Leftrightarrow\) (x - 1)[x(x - 1) - 2] = 0
\(\Leftrightarrow\) (x - 1)(x2 - x - 2) = 0
\(\Leftrightarrow\) (x - 1)(x2 - 2x + x - 2) = 0
\(\Leftrightarrow\) (x - 1)[x(x - 2) + (x - 2)] = 0
\(\Leftrightarrow\) (x - 1)(x - 2)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\end{matrix}\right.\)
Vậy S = {1; 2; -1}
x3 - 7x + 6 = 0 (3)
\(\Leftrightarrow\) x3 - x - 6x + 6 = 0
\(\Leftrightarrow\) x(x2 - 1) - 6(x - 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) - 6(x - 1) = 0
\(\Leftrightarrow\) (x - 1)[x(x + 1) - 6] = 0
\(\Leftrightarrow\) (x - 1)(x2 + x - 6) = 0
\(\Leftrightarrow\) (x - 1)(x2 + x + \(\frac{1}{4}\) - \(\frac{25}{4}\)) = 0
\(\Leftrightarrow\) (x - 1)[(x + \(\frac{1}{2}\))2 - \(\frac{25}{4}\)] = 0
\(\Leftrightarrow\) (x - 1)(x + \(\frac{1}{2}\) - \(\frac{5}{2}\))(x + \(\frac{1}{2}\) + \(\frac{5}{2}\)) = 0
\(\Leftrightarrow\) (x - 1)(x - 2)(x + 3) = 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}
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\(\text{CM vô nghiệm}\)
\(\text{a) }\left(x-2\right)^3=\left(x-2\right).\left(x^2+2x+4\right)-6\left(x-1\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6x^2+12x-6\)
\(\Leftrightarrow x^3-6x^2+12x-x^3+6x-12x=-8+8-6\)
\(\Leftrightarrow0x=-6\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)
\(\text{b) }4x^2-12x+10=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2=-1\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)
\(\text{CM vô số nghiệm}\)
\(\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^3-3x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\text{ (luôn luôn đúng)}\)
\(\text{Vậy }S\inℝ\)
a,A= x(x3-5x2+7x-3)
=x(x3-3x2-2x2+6x+x-3)
=x(x-3)(x2-2x+1)
=x(x-3)(x-1)2
vi (x-1)2>=0
=>Để A <0 thì x(x-3)<0
TH1:x>0 va x-3<0
x>0 va x<3
=> 0<x<3
TH2 :x<0 va x-3>0
x<0 và x>3( loại vỉ 2 dk trái ngược nhau )
Vay 0<x<3 thi thoa man....... .........
Phần b tương tự
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí