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16 tháng 7 2020

2x3 - 4x2 + 2x = 0

<=> x( 2x2 - 4x + 2 ) = 0

<=> x = 0 hoặc 2x2 - 4x + 2 = 0

Xét 2x2 - 4x + 2 = 0

\(\Delta=b^2-4ac=\left(-4\right)^2-4\cdot2\cdot2=0\)

\(\Delta=0\)nên phương trình đã cho có nghiệm kép 

\(x_1=x_2=\frac{-b}{2a}=\frac{-\left(-4\right)}{2\cdot2}=\frac{4}{4}=1\)

Vậy \(S=\left\{0;1\right\}\)

16 tháng 7 2020

Trả lời:

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

\(\Leftrightarrow2x.\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow2x.\left(x-1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\\left(x-1\right)^2=0\end{cases}}\)  \(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

                                               \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(x\in\left\{0,1\right\}\)

3 tháng 8 2019

\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)

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

\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)

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

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

d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)

e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)

g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

14 tháng 9 2019

a) \(x^3+2x^2+2x+1=0\)

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

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

\(TH1:x+1=0\Leftrightarrow x=-1\)

\(TH2:x^2+x+1=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)

Mà \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2

Vậy x = 1

14 tháng 9 2019

Câu a), x = -1 nha, kết luận nhầm

b) \(x^3-4x^2+12x-27=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+9\right)-4x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-7x+9\right)=0\)

\(TH1:x-3=0\Leftrightarrow x=3\)

\(TH2:x^2-7x+9=0\)

\(\cdot\Delta=\left(-7\right)^2-4.9=13\)

Vậy pt của TH2 có 2 nghiệm phân biệt

\(x_1=\frac{7+\sqrt{13}}{2}\);\(x_2=\frac{7-\sqrt{13}}{2}\)

3 tháng 10 2017

a. \(5x^2\left(2x-3\right)+\left(2x^2+3x+3\right)\left(3-2x\right)=6x^3-9x^2\Leftrightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)=3x^2\left(2x-3\right)\Leftrightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)-3x^2\left(2x-3\right)=0\Leftrightarrow\left[5x^2-\left(2x^2+3x+3\right)-3x^2\right]\left(2x-3\right)=0\Leftrightarrow\left(5x^2-2x^2-3x-3-3x^2\right)\left(2x-3\right)=0\Leftrightarrow\left(2x-3\right)\left(-3x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)

b. \(\left(4x^2+2x\right)\left(x^2-x\right)+\left(4x^2+6\right)\left(x-x^2\right)=0\Leftrightarrow\left(4x^2+2x\right)\left(x^2-x\right)-\left(4x^2+6\right)\left(x^2-x\right)=0\Leftrightarrow\left(x^2-x\right)\left[\left(4x^2+2x\right)-\left(4x^2+6\right)\right]=0\Leftrightarrow\left(x^2-x\right)\left(4x^2+2x-4x^2-6\right)=0\Leftrightarrow x\left(2x-6\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=6\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=3\\x=1\end{matrix}\right.\)

5 tháng 12 2017

1)⇔x2+1x-3x+3=0

⇔x(x+1)-3(x+1)=0

⇔(x+1)(x-3)=0

⇔x+1=0 hoặc x-3=0

⇔x=-1 hoặc x=3

5 tháng 12 2017

4)⇔x(1+5x)=0

⇔x=0 hoặc 1+5x=0

⇔x=0 hoặc 5x=-1

⇔x=0 hoặc x=-0.2

12 tháng 7 2019

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

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

\(\Leftrightarrow-5\left(4x+3\right)=0\)

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

12 tháng 7 2019

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)