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a) \(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Leftrightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Leftrightarrow2x+4=0\Leftrightarrow2x=-4\Leftrightarrow x=\dfrac{-4}{2}=-2\) vậy \(x=-2\)
\(a.\)
\(\dfrac{2x+1}{x^2-2x+1}-\dfrac{2x+3}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{2x+1}{x^2-2x+1}=\dfrac{2x+3}{x^2-1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-1\right)=\left(2x+3\right)\left(x^2-2x+1\right)\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x-1\right)=\left(2x+3\right)\left(x-1\right)^2\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)=\left(2x+3\right)\left(x-1\right)\)
\(\Rightarrow2x^2+2x+x+1=2x^2-2x+3x-3\)
\(\Rightarrow2x^2+2x+x+1-2x^2+2x-3x+3=0\)
\(\Rightarrow2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
a: =>2x^2+6x-5x-15+a+15 chia hết cho x+3
=>a+15=0
=>a=-15
b: \(\Leftrightarrow2x^3+2x-3x^2-3+5⋮x^2+1\)
=>\(x^2+1\in\left\{1;5\right\}\)
hay \(x\in\left\{0;2;-2\right\}\)
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.x(x+1)(x-1)(x+2)
b.(x+3)(x+1,5+\(\sqrt{7,25}\))(x+1,5-\(\sqrt{7,25}\))
c.(x+1)(x2-x+1)
a) Do đa thức chia có bậc là 3 , đa thức bị chia có bậc 2 nên thương sẽ có bậc 1
Ta có : (x3+ ax2 + 5x +3) = (x2+ 2x + 3)( x + d)
(x3+ ax2 + 5x +3) = x3 + dx2 + 2x2 + 2dx + 3x + 3d
(x3+ ax2 + 5x +3) = x3 + x2( d + 2) + x( 2d + 3) + 3d
Đồng nhất hệ số , ta có :
d + 2 = a --> a = 1 + 2 = 3
2d + 3 = 5 --> 2.1 + 3 = 5
3d = 3 --> d = 1
Vậy , a = 3 thỏa mãn điều kiện đề bài
b) Tẹo tớ gửi nha
a) \(\dfrac{6x-1}{2-x}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=x\left(3x-2\right)+1\)
\(\Leftrightarrow x-6x^2+2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(3x^2-25x-6-3x^2+2x-1=0\\ \Leftrightarrow-23x-7=0\\ \Leftrightarrow x=\dfrac{7}{-23}\)
vậy phương trình có tập nghiệm là S={\(\dfrac{7}{-23}\)}
b)\(\dfrac{x+2}{x-2}-\dfrac{2}{x\left(x-2\right)}=\dfrac{1}{x}\) (ĐKXĐ: \(x\ne2;0\) )
\(\Leftrightarrow\dfrac{x\left(x+2\right)-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-2=x-2\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-1}
\(2x^2+y^2+9=2xy+6x\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y=3\\x=3\end{matrix}\right.\)
Vậy.........