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4 tháng 2 2017

a) x3+4x2+x-6=0

<=> x3+x2-2x+3x2+3x-6=0

<=>x(x2+x-2)+3(x2+x-2)=0

<=>(x+3)(x2+x-2)=0

<=>(x+3)(x2+2x-x-2)=0

<=>(x+3)[x(x+2)-(x+2)]=0

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

=> x+3=0 hay

x-1=0 hay

x+2=0

<=> x=-3 hay x=1 hay x=-2

4 tháng 2 2017

b)x3-3x2+4=0

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

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

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

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

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

5 tháng 11 2017

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

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)

\(b,x^3-5x^2+4x-20=0\)

\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)

\(\Rightarrow x=5\)

5 tháng 11 2017

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

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

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

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

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

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

Vậy x=0; x=4

b) \(x^3-5x^2+4x-20=0\)

\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)

Vậy x=5

19 tháng 10 2020

a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)

\(\Leftrightarrow x=1\)

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

c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)

d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)

e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)

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

\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)

19 tháng 10 2020

a) x( x + 1 ) - x( x - 5 ) = 6

⇔ x2 + x - x2 + 5x = 6

⇔ 6x = 6

⇔ x = 1

b) 4x2 - 4x + 1 = 0

⇔ ( 2x - 1 )2 = 0

⇔ 2x - 1 = 0

⇔ x = 1/2

c) x2 - 1/4 = 0

⇔ ( x - 1/2 )( x + 1/2 ) = 0

⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)

d) 5x2 = 20x

⇔ 5x2 - 20x = 0

⇔ 5x( x - 4 ) = 0

⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) 4x2 - 9 - x( 2x - 3 ) = 0

⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0

⇔ ( 2x - 3 )( 2x + 3 - x ) = 0

⇔ ( 2x - 3 )( x + 3 ) = 0

⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)

f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )

⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0

⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0

⇔ ( 2x - 5 )(-2) = 0

⇔ 2x - 5 = 0

⇔ x = 5/2

6 tháng 9 2018

\(a,\left(x-2\right)^2=4x^2+4x+1\)

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

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

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

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

\(\Rightarrow\left(-x-3\right)\left(3x-1\right)=0\)

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

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

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

Vậy \(x\in\left\{-3;\dfrac{1}{3}\right\}\)

\(b,4x^3-4x^2+9-9x=0\)

\(\Rightarrow4x^2\left(x-1\right)+9\left(1-x\right)=0\)

\(\Rightarrow4x^2\left(x-1\right)+9\left(x-1\right)=0\)

\(\left(4x^2+9\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x^2+9=0\\x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x^2=-9\\x=1\end{matrix}\right.\)

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

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

Vậy \(x\in\left\{\sqrt{-\dfrac{3}{2}};-\sqrt{-\dfrac{3}{2}};1\right\}\)

25 tháng 10 2021

\(a,x^2-5x\)

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

\(b,5x\left(x+5\right)+4x+20\)

\(=5x\left(x+5\right)+4\left(x+5\right)\)

\(=\left(5x+4\right)\left(x+5\right)\)

\(c,7x\left(2x-1\right)-4x+2\)

\(=7x\left(2x-1\right)-2\left(2x-1\right)\)

\(=\left(7x-2\right)-\left(2x-1\right)\)

25 tháng 10 2021

\(d,x^2-16+2\left(x+4\right)\)

\(=x^2-16+2x+8\)

\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) ) 

\(e,x^2-10x+9\)

\(=x^2-x-9x+9\)

\(=x\left(x-1\right)-9\left(x-1\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé ) 

\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)

\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)

\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)

Vậy ... 

7 tháng 8 2017

1,x(2x-7)-4x-14=x(2x-7)-2(2x-7)=(2x-7)(x-2)

15 tháng 7 2017
  1. Tập xác định của phương trình

  2. 2

    Rút gọn thừa số chung

  3. 3

    Biệt thức

  4. 4

    Biệt thức

  5. 5

    Nghiệm

16 tháng 7 2017

phaỉ giải rõ ra bạn nhé !