\(x^6+2x^3+1=0\)

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

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

\(\Leftrightarrow x^3=\left(-1\right)^3\)

\(\Leftrightarrow x=-1\)

___________

\(x\left(x-5\right)=4x-20\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

_____________

\(x^4-2x^2=8-4x^2\)

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

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

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

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

_______________

\(\left(x^3-x^2\right)-4x^2+8x-4\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

x ∈ N*

\(\Rightarrow\left(4x^2-4x+1\right)-2=0\\ \Rightarrow\left(2x-1\right)^2=2\\ \Rightarrow\left[{}\begin{matrix}2x-1=\sqrt{2}\\2x-1=-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{2}+1}{2}\\x=\dfrac{1-\sqrt{2}}{2}\end{matrix}\right.\)

`a)16x^2-24x+9=25`

`<=>(4x-3)^2=25`

`+)4x-3=5`

`<=>4x=8<=>x=2`

`+)4x-3=-5`

`<=>4x=-2`

`<=>x=-1/2`

`b)x^2+10x+9=0`

`<=>x^2+x+9x+9=0`

`<=>x(x+1)+9(x+1)=0`

`<=>(x+1)(x+9)=0`

`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2+2x-6x-12=0`

`<=>x(x+2)-6(x+2)=0`

`<=>(x+2)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2+x-6x-6=0`

`<=>x(x+1)-6(x+1)=0`

`<=>(x+1)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

`e)4x^2-3x-1=0`

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

`<=>4x(x-1)+(x-1)=0`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\) 

`f)x^4+4x^2-5=0`

`<=>x^4-x^2+5x^2-5=0`

`<=>x^2(x^2-1)+5(x^2-1)=0`

`<=>(x^2-1)(x^2+5)=0`

Vì `x^2+5>=5>0`

`=>x^2-1=0<=>x^2=1`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

tham khảo: https://hoc24.vn/cau-hoi/.2256230161739

a) ⇔ \(4x^2+4x-x-1=0\)

⇔ \(4x^2+3x-1=0\)

⇔ \(4x(x+1)-(x+1)=0\)

⇔ \((x+1)(4x-1)=0\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

b) \(x^3-4x^2+4x=0\)

⇔ \(x^2(x-2)-2x(x-2)=0\)

⇔ \((x-2)(x^2-2x)=0\)

⇒ \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy...

c) \(x^2-3x+2=0\)

⇔ \(x(x-2)-(x-2)=0\)

⇔ \((x-1)(x-2)=0\)

⇒ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy...

Cho mk xin yêu cầu của bài được ko vậy ???

a. 2x²-5x+1=0

△= b² - 4ac = (-5)² - 4*2*1 = 17 ⇒√△ = √17

\(\Rightarrow x_1=\frac{5+\sqrt{17}}{4};x_2=\frac{5-\sqrt{17}}{4}\)

Vậy .... S={\(\frac{5\pm\sqrt{17}}{4}\)}

b. 4x² +4x+1=0

⇔(2x+1)² = 0 ⇔ x=\(\frac{-1}{2}\)

c. -3x² +2x+8=0

△' = b'² - ac = 1² - (-3)*8 = 25 ⇒√△ = 5

\(\Rightarrow x_1=\frac{-1+5}{-3}=-\frac{4}{3};x_2=\frac{-1-5}{-3}=2\)

Vậy... S={-\(\frac{4}{3}\);2}

d. 5x2 6x1=0 (thiếu dấu nên mk chưa giải được)

e. -3x²+ 14x - 8=0

△' = b'² - ac = 7² - (-3)*(-8) = 25 ⇒ √△ = 5

⇒\(x_1=\frac{-7+5}{-3}=\frac{2}{3};x_2=\frac{-7-5}{-3}=4\)

Vậy .... S={\(\frac{2}{3};4\)}

g. -7x² +4x-3=0

△' = b'² - ac = 2² - (-7)*(-3) = -17<0

Vậy pt vô nghiệm , S=∅

\(a,\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ b,4x^2-1=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(c,x^2-4x+3=0\\ \Leftrightarrow x^2-3x-x+3=0\\ \Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,9x^2-6x+1=0\\ \Leftrightarrow\left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow x=\dfrac{1}{3}\)

\(a\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\)

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)