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

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

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

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

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

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

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

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

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

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 \(x=-3\)

pt quá vĩ đại =.= cx trên OLM lun 

câu a biến đổi to lắm

\(\Leftrightarrow-\left(12x\sqrt{6x-1}-2\sqrt{6x-1}-2x^3-9x^2+6x-8\right)=0\)rồi sao nx 

cái này ra nghiệm là 

\(2-\sqrt{2}\)và\(\sqrt{2}+2\)

1: \(\Leftrightarrow x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+2008\right)=0\)

hay \(x\in\varnothing\)

2: \(x^4+x^2+6x-8=0\)

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

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

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

hay \(x\in\left\{1;-2\right\}\)

=>x(x*căn 2+6)=0

=>x=0