Thử sức những bạn giỏi toán .
Giải phương trình :
a ) \(x^4+2x^3-4x^2-5x-6=0\)
b ) \(6x^4-5x^3-38x^2-5x+6=0\)
ĐÚNG tặng 1SP
Phùng Khánh Linh Nhã Doanh v...v
b) 6x4 - 5x3 - 38x2 - 5x + 6 = 0
⇔ x2( 6x2 - 5x - 38 -\(\dfrac{5}{x}\) + \(\dfrac{6}{x^2}\) ) = 0
⇔ 6x2 - 5x - 38 - \(\dfrac{5}{x}\) + \(\dfrac{6}{x^2}\) = 0
⇔ 6( x2 + \(\dfrac{1}{x^2}\)) - 5( x + \(\dfrac{1}{x}\)) - 38 = 0
Đặt : x + \(\dfrac{1}{x}\) = y ⇒ x2 + \(\dfrac{1}{x^2}\) = y2 - 2
Ta có : 6( y2 - 2) - 5y - 38 = 0
⇔ 6y2 - 12 - 5y - 38 = 0
⇔ 6y2 - 5y - 50 = 0
⇔ 6y2 + 15y - 20y - 50 = 0
⇔ 2y( 3y - 10 ) + 5( 3y - 10 ) = 0
⇔ ( 3y - 10 )( 2y + 5) = 0
⇔ y = \(\dfrac{10}{3}\) hoặc : y = \(\dfrac{-5}{2}\)
*) Với : y = \(\dfrac{10}{3}\) , ta có :
x + \(\dfrac{1}{x}\) = \(\dfrac{10}{3}\)
⇔ \(\dfrac{x^2+1}{x}\) = \(\dfrac{10}{3}\) ( x # 0)
⇔ 3x2 - x - 9x + 3 = 0
⇔ x( 3x - 1) - 3( 3x - 1) = 0
⇔ ( 3x - 1)( x - 3) = 0
⇔ x = \(\dfrac{1}{3}\) ( TM ) hoặc : x = 3 ( TM)
*) Với : y = \(\dfrac{-5}{2}\) , ta có :
x + \(\dfrac{1}{x}\) = \(\dfrac{-5}{2}\)
⇔ \(\dfrac{x^2+1}{x}\) = \(\dfrac{-5}{2}\) ( x # 0)
⇔ 2x2 + 2 + 5x = 0
⇔ 2x2 + x + 4x + 2 = 0
⇔ x( 2x + 1) + 2( 2x + 1) = 0
⇔ x = - 2 hoặc : x = \(\dfrac{-1}{2}\)
a/ \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4+x^3+x^2-2x^3-2x^2-2x+3x^3+3x^2+3x-6x^2-6x-6=0\)
\(\Leftrightarrow x^2\left(x^2+x+1\right)-2x\left(x^2+x+1\right)+3x\left(x^2+x+1\right)-6\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x-2\right)\left(x+3\right)=0\)
Vì \(x^2+x+1>0\forall x\Rightarrow\) vô nghiệm
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy pt có 2 nghiệm....
b/ \(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4+3x^3-2x^3-18x^3-9x^2+6x^2+3x-12x^3-6x^2+4x^2+2x+36x^2+18x-12-6=0\)
\(\Leftrightarrow3x^3\left(2x+1\right)-x^2\left(2x+1\right)-9x^2\left(2x+1\right)+3x\left(2x+1\right)-6x^2\left(2x+1\right)+2x\left(2x+1\right)+18x\left(2x+1\right)-6\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x^3-x^2-9x^2+3x-6x^2+2x+18x-6\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[x^2\left(3x-1\right)-3x\left(3x-1\right)-2x\left(3x-1\right)+6\left(3x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\3x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\\x=2\\x=3\end{matrix}\right.\)
a) Ko chép đề nha
a) x4 - 2x3 + 4x3 - 8x2 + 4x2 - 8x + 3x - 6 = 0
⇔ x3( x - 2) + 4x2( x - 2) + 4x( x - 2) + 3(x - 2) = 0
⇔ ( x - 2)( x3 + 4x2 + 4x + 3) = 0
⇔ ( x - 2)( x3 + 3x2 + x2 + 3x + x + 3) = 0
⇔ ( x - 2)[ x2( x + 3) + x( x + 3 ) + ( x + 3) ] = 0
⇔ ( x - 2)( x + 3)( x2 + x + 1 ) = 0
Do : x2 + x + 1 = \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) > 0 ∀x
⇔ x = 2 hoặc : x = - 3
KL...
