Ôn tập phép nhân và phép chia đa thức

\(\dfrac{12x}{5y^2}:\dfrac{3x^2}{10y^3}=\dfrac{12x}{3x^2}\cdot\dfrac{10y^3}{5y^2}=\dfrac{4}{x}\cdot2y=\dfrac{8y}{x}\)

\(a,\left(3x-1\right)^2+1-3x=0\)
\(\Rightarrow\left(3x-1\right)\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(3x-1-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)       Vậy \(x\in\left\{\dfrac{1}{3};\dfrac{2}{3}\right\}\)

\(VT=\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-x^2+2x-1-1}\)

         \(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{y\left(x^2+2-2x\right)-\left(x^2-2x+2\right)}\)

         \(=\dfrac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{\left(y-1\right)\left(x^2-2x+2\right)}=\dfrac{x^2+2x+2}{y-1}\)

\(\Rightarrow VT=VP\)

\(VT=\dfrac{3n-2-m\left(3n-2\right)}{1-3.m.1+3.m^2.1-m^3}=\dfrac{\left(3n-2\right)\left(1-m\right)}{\left(1-m\right)^3}\)

  \(=\dfrac{3n-2}{\left(1-m\right)^2}\Rightarrow VT=VP\)

Bài 10:

a: DKXĐ: x<>1

\(Q=\dfrac{3+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-1}{x^2+x+1}\)

b: x^2=1

=>x=1(loại) hoặc x=-1(nhận)

Khi x=-1 thì \(Q=\dfrac{-1}{\left(-1\right)^2+\left(-1\right)+1}=\dfrac{-1}{1}=-1\)

c: x^2+x+1=(x+1/2)^2+3/4>0

=>Q<0

a: Khi x=-3 thì Q=(-3+1)/(-3)=-2/-3=2/3

b: \(P=\dfrac{x^2-2+x}{x\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x-1\right)}{x\left(x+2\right)}=\dfrac{x-1}{x}\)

c: P/Q=(x-1)/x:(x+1)/x=x-1/x+1

Để P/Q=5/2 thì x-1/x+1=5/2

=>5x+5=2x-2

=>3x=-7

=>x=-7/3

=>2(x^2+2x+1)+4x^2+4x+1-6(x-1)(x+1)=10

=>2x^2+4x+2+4x^2+4x+1-6x^2+6=10

=>8x+9=10

=>8x=1

=>x=1/8

a: =16-(x^2-4xy+4y^2)

=16-(x-2y)^2

=(4-x+2y)(4+x-2y)

b: =(x-3)(x-4)

a: \(=9x^2-12x+4+5\left(1-x^2\right)\)

\(=9x^2-12x+4+5-5x^2=4x^2-12x+9\)

b: \(=\dfrac{x^2+x-6-\left(2x-1\right)\left(x+2\right)-x+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2-3-2x^2-4x+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-x^2-3x+1}{\left(x-2\right)\left(x+2\right)}\)

c: \(=\dfrac{6x^3+3x^2-10x^2-5x+6x+3}{2x+1}=3x^2-5x+2\)

`P=2x^2 -4x+5`

`P=2(x^2 -2x +5/2)`

`P=2(x^2 -2.x.1 +1 -1 +5/2)`

`P=2[(x-1)^2 +3/2]≥3` với mọi `x`

Vì `(x-1)^2 ≥0` với mọi `x`

Dấu "=" xảy ra khi `x-1=0=>x=1`

Vậy \(P_{min}\) là:`3` khi `x=1`

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

Do \(\left(x-1\right)^2\ge0;\forall x\)

\(\Rightarrow P\ge3;\forall x\)

\(P_{min}=3\) khi \(x=1\)