ta có:

pt trên \(< =>x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)

\(< =>\left[\left(x^2+6x\right)+1\right]^2=\left(2x+1\right)^2.\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+36x^2+2.\left(x^2+6x\right)+1=\left(4x^2+4x+1\right)\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+38x^2+12x+1=\)

\(4x^4+8x^3+12x^2+4x^3+8x^2+12x+x^2+2x+3\)

\(=4x^4+12x^3+21x^2+14x+3\)

\(< =>-3x^4+17x^2-2x-2=0\)

\(< =>-\left(x^2+2x-1\right)\left(3x^2-6x+2\right)=0\)

đến đây dễ rùi bạn tự giải nhé 

\(\text{Đ}K:x^2+2x+3\ge0\\ x^2+6x+1=\left(2x+1\right)\cdot\sqrt{x^2+2x+3}\\ \Leftrightarrow x^2+2x+3+4x+2=\left(2x+1\right)\cdot\sqrt{x^2+2x+3+4}\)

\(\text{ Đặt }\)\(m=\sqrt{x^2+2x+3};n=2x+1\) \(\text{ phương trình trở thành :}\)

\(m^2+2n=mn+4\\ \Leftrightarrow m^2-4-mn+2n=0\\ \Leftrightarrow\left(m-2\right)\left(m+2\right)-n\left(m-2\right)=0\\ \Leftrightarrow\left(m-2\right)\left(m-n-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\\m-n=-2\end{matrix}\right.\)

`\text{ Với}` \(m=2\\ \Leftrightarrow\sqrt{x^2+2x+3}=2\Leftrightarrow x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(N\right)\\x=-\sqrt{2}-1\left(N\right)\end{matrix}\right.\)

`\text{Với}`\(m-n=-2\Leftrightarrow\sqrt{x^2+2x+3}-\left(2x+1\right)=-2\\ \Leftrightarrow\sqrt{x^2+2x+3}=-2+2x+1=2x-1\\ \Leftrightarrow x^2+2x+3=4x^2-4x+1\\ \Leftrightarrow3x^2-6x-2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{15}}{3}\left(N\right)\\x=\dfrac{3-\sqrt{15}}{3}\left(L\right)\end{matrix}\right.\)

2x² - 6x + 1 = 0 

Có: \(\Delta=\left(-6\right)^2-4.2.1=28\Rightarrow\sqrt{\Delta}=2\sqrt{7}\)

\(\Rightarrow x_1=\frac{6+2\sqrt{7}}{4}=\frac{3+\sqrt{7}}{2}\) hoặc \(x_1=\frac{3-\sqrt{7}}{2}\)

Vậy \(x=\left\{\frac{3+\sqrt{7}}{2};\frac{3-\sqrt{7}}{2}\right\}\)

(-6)²-4(2.1)=28

\(x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{6\pm\sqrt{28}}{4}\)

x₁=\(-\frac{\sqrt{7}-3}{2}\);x₂=\(\frac{\sqrt{7}+3}{2}\)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

\(\Delta^'=\left(-3\right)^2-2.1=7\)

Nghiệm của phương trình : \(\orbr{\begin{cases}x=\frac{3-\sqrt{7}}{2}\\x=\frac{3+\sqrt{7}}{2}\end{cases}}\)

Mình giải cách khác !!!

\(2x^2-6x+1=0\\ \Rightarrow2\left(x^2-3x\right)+1=0\\ \Rightarrow2\left(x^2-2.\frac{3}{2}.x+\frac{3}{2}.\frac{3}{2}\right)+1-\frac{9}{2}=0\\ \)

\(\Rightarrow2.\left(x-\frac{3}{2}\right)^2-\frac{7}{2}=0\\ \Rightarrow\left(x-\frac{3}{2}\right)^2=\frac{7}{2}\\ \Rightarrow\hept{\begin{cases}x-\frac{3}{2}=\sqrt{\frac{7}{2}}\\x-\frac{3}{2}=-\sqrt{\frac{7}{2}}\end{cases}}\)

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

tl

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

^HT^

Ta thấy x=0 không là nghiệm của phương trình

chia cả 2 vế cho x^2 ta được:

PT <=> x^2-3x-6+3/x+1/(x^2)=0

       <=> (x^2-2+1/(x^2))-3(x-1/x)-4=0

      <=> (x-1/x)^2-3(x-1/x)-4=0

Đặt x-1/x=y

PT <=> y^2-3y-4=0

     <=> y=-4 hoặc y=1

Tại y=-4 , ta có x+1/x+4=0

                       <=> x^2+4x+1=0

                       <=> x=-2+ √3 hoăc x=-2-  √ 3

Tại y=1 ta có x^2-x-1=0

                 <=> x=(1- √  5)/2 hoặc x=(1+  √5)/2