Tìm đa thức \(f\left(x\right)\), biết rằng \(f\left(x\right)\) chia cho \(x-3\) thì dư 2, \(f\left(x\right)\) chia cho \(x+4\) thì dư \(9\), còn \(f\left(x\right)\) chia hết cho \(x^2+x-12\) thì được thương là \(x^2+3\) và còn dư.
Lời giải:
Vì $f(x)$ chia $x-3$ dư $2$, $f(x)$ chia $x+4$ dư $9$ nên $f(3)=2; f(-4)=9$
Giả sử $f(x)$ chia $x^2+x-12$ được đa thức dư là $ax+b$
Khi đó: $f(x)=(x^2+x-12)(x^2+3)+ax+b$
$f(3)=(3^2+3-12)(3^2+3)+3a+b$
$\Leftrightarrow 2=3a+b(1)$
$f(-4)=[(-4)^2-4-12][(-4)^2+3)]-4a+b$
$\Leftrightarrow 9=-4a+b(2)$
Từ $(1);(2)\Rightarrow a=-1; b=5$
$f(x)=(x^2+x-12)(x^2+3)-x+5=x^4+x^3-9x^2+2x-31$
Rút gọn biểu thức
Q=\(\left(\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right):\dfrac{x+1}{2x^2+y+2}\)
R=\(\left(5-2x^2\right):\left(\dfrac{1+x+x^2}{2x+x^2}+2-\dfrac{1-x+x^2}{2x-x^2}\right)\)
Các bạn làm gấp nha bài nào cũng được mình đang cần gấp làm rồi mình tick cho
Mog mọi giúp cho mình càng sớm càng tốt
Điền vào chỗ trống:
\(\dfrac{x}{x+1}\):\(\dfrac{x+2}{x+1}\):\(\dfrac{x+3}{x+2}\): .......... =\(\dfrac{x}{x+6}\)
\(\dfrac{x}{x+1}:\dfrac{x+2}{x+1}:\dfrac{x+3}{x+2}:\) ________ \(=\dfrac{x}{x+6}\)
\(\Rightarrow\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\) ________\(=\dfrac{x}{x+6}\)
\(\Rightarrow\dfrac{x}{x+3}\)______ = \(\dfrac{x}{x+6}\)
\(\Rightarrow\) _______ = \(\dfrac{x+3}{x+6}\)
Bài 1:
a, x\(^2\)(x - 2x\(^3\)) b,(x\(^2\) + 1)(5 - x) c,(x - 2)(x\(^2\) + 3x - 4) d,(x - 2y)\(^2\) e,(2x\(^2\) + 3)\(^2\) f,(x - 2)(x\(^2\) + 2x + 4) g,(2x-1)\(^3\)
b: \(x^2\left(x-2x^3\right)=x^3-2x^5\)
b: \(\left(x^2+1\right)\left(5-x\right)\)
\(=5x^2-x^3+5-x\)
c: \(\left(x-2\right)\left(x^2+3x-4\right)\)
\(=x^3+3x^2-4x-2x^2-6x+8\)
\(=x^3+x^2-10x+8\)
d: \(\left(x-2y\right)^2=x^2-4xy+4y^2\)
x^2-2xy+y2+2x-2y
Giúp em vs ạ
PTĐTTNT?
\(x^2-2xy+y^2+2x-2y\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)^2+2\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)+2\right]\)
\(=\left(x-y\right)\left(x-y+2\right)\)
Thực hiện phép tính:
a) \(\dfrac{2x ^2-20x+50}{3x+3}.\dfrac{x^2-1}{4\left(x-5\right)^3}\)
b)\(\dfrac{x^{ }^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
Thực hiện phép tính:
b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x^2+x}{5x^2-10x+5}.\dfrac{5x-5}{3x+3}\)
\(=\dfrac{\left(x^2+x\right)\left(5x-5\right)}{\left(5x^2-10x+5\right)\left(3x+3\right)}\)
\(=\dfrac{x\left(x+1\right)5\left(x-1\right)}{5\left(x^2-2x+1\right)3\left(x+1\right)}\)
\(=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x-1\right)}\)
\(\dfrac{x^2-y^2}{6x^2y}\):\(\dfrac{x+y}{3xy}\)
Mấy bạn giúp mình đi :3
\(\dfrac{x^2-y^2}{6x^2y}:\dfrac{x+y}{3xy}\)
\(=\dfrac{x^2-y^2}{6x^2y}.\dfrac{3xy}{x+y}\)
\(=\dfrac{3xy\left(x^2-y^2\right)}{6x^2y\left(x+y\right)}\)
\(=\dfrac{3xy\left(x-y\right)\left(x+y\right)}{6x^2y\left(x+y\right)}\)
\(=\dfrac{x-y}{2x}\)
rút gọn:
\(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right):\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
Bài này nhân chứ sao lại chia :v Có trong SBT mà :v
\(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right).\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{2\left(x^2+4\right)-x\left(x^2+4\right)}\right].\dfrac{x^2-x-2}{x}\)
\(=\left[\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+3\right)}\right].\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(2-x\right)-4x^2}{2\left(2-x\right)\left(x^2+4\right)}.\dfrac{x^2+x-2x-2}{x^2}\)
\(=\dfrac{-x\left(x^2+4\right)}{2\left(2-x\right)\left(x^2+4\right)}.\dfrac{\left(x+1\right)\left(x-2\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
