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Bài 2:
a: \(=\dfrac{4x^2+3-19}{x-2}=\dfrac{4x^2-16}{x-2}=\dfrac{4\left(x-2\right)\left(x+2\right)}{x-2}=4x+8\)
b: \(=\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2}{x+2y}-\dfrac{1}{x-2y}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x-4y-x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{x-6y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=\left(4x^2+2x-18-4x^2-2x\right)\left(4x^2+2x-18+4x^2+2x\right)\)
\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=-18\cdot\left(8x^2+4x-18\right)\)
\(\Leftrightarrow M=-18\cdot2=-36\)
Ta có
4 x 2 + 2 x - 18 2 - 4 x 2 + 2 x 2 = 4 x 2 + 2 x - 18 + 4 x 2 + 2 x 4 x 2 + 2 x - 18 - 4 x 2 - 2 x = 8 x 2 + 4 x - 18 - 18 = 2 4 x 2 + 2 x - 9 - 18 = - 36 4 x 2 + 2 x - 9 ⇒ m = - 36
Đáp án cần chọn là: C
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Ohhh, tui hiểu r.
x2 - x + \(\dfrac{1}{4}\)
⇔ x2 - 2.\(\dfrac{1}{2}\).x + \(\left(\dfrac{1}{2}\right)^2\)
⇔ \(\left(x^2-\dfrac{1}{2}\right)^2\)
1: Ta có: \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
4x2 - 9
= ( 2x)2 - 32
= ( 2x - 3 )( 2x + 3 )
Chak v :p
\(4x^2-9\)
\(=\left(2x\right)^2-3^2\)
\(=\left(2x-3\right)\left(2x+3\right)\)