b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

\(\Leftrightarrow\dfrac{2}{-x^2+6x-8}=\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\\ \Leftrightarrow\left\{{}\begin{matrix}2=\left(-x^2+6x-8\right)\left(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\right)\\-x^2+6x-8\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2=-2x^2+4x+2\\-x^2+6x-8\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\-x^2+6x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\-x^2+6x-8\ne0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\-x^2+6x-8\ne\end{matrix}\right.\end{matrix}\right.\\\Rightarrow x=0\)

\(x^4-6x^3+7x^2+6x-8=0\)

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

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

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

\(\Leftrightarrow\left(x-4\right)\left[x^2\left(x-2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{-1;1;2;4\right\}\)

Vậy S={-1;1;2;4}

g: =>(x-1)(x-2)=0

=>x=1 hoặc x=2

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

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

=>x=1 hoặc x=-2

g) x^2 - 3x + 2 = 0

<=> x^2 - 2x-x+2 =0

<=> x=1 hoặc x = 2

..tự kết luận

i)x^4 + x^2 + 6x - 8=0

<=> x^4 + 2x^3 - 2x^3 - 4x^2 + 5x^2 + 10x - 4x - 8 = 0

<=> x^3(x + 2) - 2x^2(x+2) + 5x(x+2) - 4(x+2) = 0

<=> (x^3 - 2x^2 +5x -4)(x+2)=0

<=> (x^3 - x^2 -x^2 +x + 4x - 4)(x+2) = 0

<=>(x^2(x-1) - x(x-1) + 4(x-1) )(x+2) = 0

<=> (x^2-x+4)(x-1)(x+2)=0

<=> x = 1 hoặc x +-2 hoặc x^2 - x+4=0

                                     <=>x^2 - x+ 1/4 - 1/4 +4=0

                                      <=>(x-1/2)^2 +15/4=0

                                     <=>(x-1/2)^2=-15/4 (vô lí)

....tự kết luận

h)x^3 - 8x^2 + 21x - 18 = 0

<=> x^3 - 2x^2 - 6x^2 + 12x + 9x - 18 = 0

<=> x^2(x-2) -6x(x-2) + 9(x-2) =0

<=>(x-3)^2(x-2)=0

<=> x=3 hoặc x =2 

...tự kết luận

g: \(x^2-3x+2=0\)

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

=>x=1 hoặc x=2

i: \(x^4+x^2+6x-8=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x+2\right)\left(x^2-2x+4\right)+x\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

=>x=1 hoặc x=-2

h) \(x^3-8x^2+21x-18=0\)

\(\Leftrightarrow x^3-2x^2+6x^2-12x+9x-18=0\)

\(\Leftrightarrow x^2\left(x-2\right)+6x\left(x-2\right)+9\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+6x+9\right)=0\)

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

\(\Leftrightarrow x-2=0\) hay \(x+3=0\)

\(\Leftrightarrow x=2\) hay \(x=-3\)

\(2\sqrt{6x-5}+\sqrt{x^2-6x+14}=x^2-4x+8\\ \Leftrightarrow2\left(\sqrt{6x-5}-5\right)+\sqrt{x^2-6x+14}-3=x^2-4x-5\)

(đk x>= 5/6)

\(\Leftrightarrow\frac{2\left(6x-5-25\right)}{\sqrt{6x-5}+5}+\frac{x^2-6x+5}{\sqrt{x^2-6x+14}+3}=\left(x+1\right)\left(x-5\right)\)

\(\Leftrightarrow\frac{12\left(x-5\right)}{\sqrt{6x-5}+5}+\frac{\left(x-1\right)\left(x-5\right)}{\sqrt{x^2-6x+14}+3}-\left(x+1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{12}{\sqrt{6x-5}+5}+\frac{x-1}{\sqrt{x^2-6x+14+3}}-x-1\right)=0\)

suy ra x = 5 ( dễ dàng chứng minh được cái ngoặc còn lại luôn dương với mọi x lớn hơn bằng 5/6 )

vậy x = 5 là nghiệm của phương trình