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

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

\(\Leftrightarrow-x^3+x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow-x^2\left(x-1\right)-7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x^2-7x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-x^2-7x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+12,25-11,25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5\right)^2-\left(\frac{3\sqrt{5}}{2}\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5-\frac{3\sqrt{5}}{2}\right)\left(x+3,5+\frac{3\sqrt{5}}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+3,5-\frac{3\sqrt{5}}{2}=0\\x+3,5+\frac{3\sqrt{5}}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3,5+\frac{3\sqrt{5}}{2}=\frac{-7+3\sqrt{5}}{2}\\x=-3,5-\frac{3\sqrt{5}}{2}=\frac{-7-3\sqrt{5}}{2}\end{matrix}\right.\)

Vậy x = \(\left\{1;\frac{-7+3\sqrt{5}}{2};\frac{-7-3\sqrt{5}}{2}\right\}\)

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

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

\(\Leftrightarrow x^3-x^2+7x^2-7x+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2-\frac{45}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2=\left(\frac{\pm\sqrt{45}}{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\pm\sqrt{45}-7}{2}\end{matrix}\right.\)

2x⁴ - 6x³ + x² + 6x - 3 = 0

=> 2x⁴ - 2x³ - 4x³ + 4x² - 3x² + 3x + 3x - 3 = 0

=> 2x³(x - 1) - 4x²(x - 1) - 3x(x - 1) + 3(x - 1) = 0

=> (x - 1)(2x³ - 4x² - 3x + 3) = 0

=> (x - 1)(2x³ + 2x² - 6x² - 6x + 3x + 3) = 0

=> (x - 1)[2x²(x + 1) - 6x(x + 1) + 3(x + 1)] = 0

=> (x - 1)(x + 1)(2x² - 6x + 3) = 0

\(\Rightarrow\left[{}\begin{matrix}x-1\\x+1\\2x^2-6x+3\end{matrix}\right.\) (2x² - 6x + 3 vô nghiệm)

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

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

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\left(2x+1\right)+\left(2x+1\right)-\sqrt[3]{1+6x}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{\sqrt{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{\sqrt{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-4}{2}+\dfrac{12}{3}=...\)

giup minh voi dang can

ai làm có thưởng 2điem

d) (x - 2)^2 = 1 

= x = 2 + 1 = 3  

c) (x^2 + 1). (x + 2011) = 0

Tim x:

a) x^2 + 2x = 0 

= \(x^2+2x=0\)

= \(x^2=0:2=0\)

b) (x - 3) + 2x^2 - 6x = 0

Rút gọn thừa số chung : 

\(2x^2-5x-3=0\)

x = \(\frac{-1}{2}\)x = 3

=\(x^2=0\)

=> x = 0 

Bạn Nguyễn Phương Trung làm đúng rồi, các bn tk cho bn ấy nha !