Ôn tập phép nhân và phép chia đa thức

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

a) Ta có: \(C=\left(x^2-1\right)\cdot\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}+1\right)\)

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

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

\(=x^2-1\)

\(A=3yz+\left(4-y-z\right)\left(y+2z\right)\)

\(A=-y^2+4y-2z^2+8z\)

\(A=-\left(y-2\right)^2-2\left(z-2\right)^2+12\le12\)

\(A_{max}=12\) khi \(\left(x;y;z\right)=\left(0;2;2\right)\)

a) (x + 2)(x² + 3x + 1)

= x.x² + x.3x + x.1 + 2.x² + 2.3x + 2.1

= x³ + 3x² + x + 2x² + 6x + 2

= x³ + 5x² + 7x + 2

b) (2x³ + 10x² + 9x + 4) : (x + 4)

= (2x³ + 8x² + 2x² + 8x + x + 4) : (x + 4)

= [(2x³ + 8x²) + (2x² + 8x) + (x + 4)] : (x + 4)

= [2x²(x + 4) + 2x(x + 4) + (x + 4)] : (x + 4)

= (x + 4)(2x² + 2x + 1) : (x + 4)

= 2x² + 2x + 1

giúp mik vs ạ

câu 1 phải ko bạn

Bài 1: 

a) Ta có: \(A=\left(x+3\right)^2+2\left(x+3\right)\left(x-3\right)-3x\left(x-3\right)\)

\(=x^2+6x+9+2\left(x^2-9\right)-3x^2+9x\)

\(=-2x^2+15x+9+2x^2-18\)

\(=15x-9\)

b) Ta có: \(\dfrac{2x}{3x-3y}-\dfrac{x^2}{x-y}\)

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

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

c) Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=0\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

(x-3)(x² + 3x + 9) - (3x-17) = x(x² - 8)

<=> x³ - 27 - 3x + 17 = x³ - 8x

<=> 5x - 10 = 0

<=> 5x = 10

<=> x = 2

Vậy ...

x² - (1-x²) -4 + 4x² = x² -1 + x² - 4 + 4x² = 6x² - 5

=   

=   

a ,\(4x^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(2x-x+3\right)\left(2x+x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\3x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Vậy 

b,\(x^2-4+\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy ...

a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)

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

=2x

b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)

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

\(=\dfrac{x+1}{3x}\)

c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)

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

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

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

a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )

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

c

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