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

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

Từ (1) ; (2) \(\Rightarrow\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) (đpcm)

b ) \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\) (3)

\(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\) (4)

Từ (3) và (4) \(\Rightarrow\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\) (đpcm)

a )\(A=2x^2-8x-10=2\left(x^2-4x-5\right)=2\left[\left(x^2-4x+4\right)-9\right]\)

\(=2\left[\left(x-2\right)^2-9\right]=2\left(x-2\right)^2-18\)

Vì \(2\left(x-2\right)^2\ge0\forall x\) nên \(A=2\left(x-2\right)^2-18\ge-18\forall x\)

Dấu "=" xảy ra <=> \(2\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy GTNN của A là - 18 tại x = 2

b ) \(B=9x-3x^2=-3\left(x^2-3x\right)=-3\left[\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}\right]\)

\(=-3\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]=-3\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

Vì \(\cdot3\left(x-\dfrac{3}{2}\right)^2\le0\forall x\) nên \(B=-3\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\le\dfrac{27}{4}\)

Dấu "=" xảy ra <=> \(-3\left(x-\dfrac{3}{2}\right)^2=0\Rightarrow x=\dfrac{3}{2}\)

Vậy GTLN của B là \(\dfrac{27}{4}\) tại x = \(\dfrac{3}{2}\)

Gỉa sử Q(x) có nghiệm

\(\Rightarrow x^2+\left(x+2\right)^2=0\)

Vì \(x^2\ge0\forall x;\left(x+2\right)^2\ge0\forall x\Rightarrow x^2+\left(x+2\right)^2\ge0\forall x\)

Để \(x^2+\left(x+2\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\\left(x+2\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\) (loại vì vô lí)

Vậy điều giả sử là sai

Từ đó suy ra \(Q\left(x\right)=x^2+\left(x+2\right)^2\) vô nghiệm

\(\dfrac{3x}{x+3}=\dfrac{3x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x^2-9x}{x^2-9}\)

\(\dfrac{x-1}{x-3}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2x-3}{x^2-9}\)

\(x^2+9=\dfrac{\left(x^2+9\right)\left(x^2-9\right)}{x^2-9}=\dfrac{x^4-81}{x^2-9}\)

\(\dfrac{1}{1-x}\cdot\dfrac{1}{1+x}\cdot\dfrac{1}{1+x^2}\cdot\dfrac{1}{1+x^4}\cdot\dfrac{1}{1+x^8}\cdot\dfrac{1}{1+x^{16}}\)

\(=\dfrac{1}{\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

\(=\dfrac{1}{\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

\(=\dfrac{1}{\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

\(=\dfrac{1}{\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

\(=\dfrac{1}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)

\(=\dfrac{1}{1-x^{32}}\)

a ) \(\dfrac{x-y}{x^3+y^3}.Q=\dfrac{x^2-2xy+y^2}{x^2-xy+y^2}\)

\(\Leftrightarrow Q=\dfrac{x^2-2xy+y^2}{x^2-xy+y^2}:\dfrac{x-y}{x^3+y^3}\)

\(\Leftrightarrow Q=\dfrac{\left(x-y\right)^2}{x^2-xy+y^2}\cdot\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x-y}\)

\(\Rightarrow Q=\left(x-y\right)\left(x+y\right)=x^2-y^2\)

Vậy \(Q=x^2-y^2\)

b ) \(\dfrac{x+y}{x^3-y^3}.Q=\dfrac{3x^2+3xy}{x^2+xy+y^2}\)

\(\Leftrightarrow Q=\dfrac{3x^2+3xy}{x^2+xy+y^2}:\dfrac{x+y}{x^3-y^3}\)

\(\Leftrightarrow Q=\dfrac{3x\left(x+y\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x+y}\)

\(\Leftrightarrow Q=3x\left(x-y\right)=3x^2-3xy\)

Vậy \(Q=3x^2-3xy\)

Gọi phân thức cần tìm là \(\dfrac{a}{b}\)

Theo đề bài ta có :

\(\dfrac{x}{x+1}:\dfrac{x+2}{x+1}:\dfrac{x+3}{x+2}:\dfrac{x+4}{x+3}:\dfrac{x+5}{x+4}:\dfrac{a}{b}=1\)

\(\Leftrightarrow\dfrac{x}{x+1}\cdot\dfrac{x+1}{x+2}\cdot\dfrac{x+2}{x+3}\cdot\dfrac{x+3}{x+4}\cdot\dfrac{x+4}{x+5}\cdot\dfrac{b}{a}=1\)

\(\Leftrightarrow\dfrac{x}{x+5}\cdot\dfrac{b}{a}=1\)

\(\Rightarrow\dfrac{b}{a}=\dfrac{x+5}{x}\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{x}{x+5}\)

Vậy phân thức cần tìm là \(\dfrac{x}{x+5}\)

a ) Gọi \(A=\dfrac{3x^2-x}{9x^2-6x+1}\)

Ta có : \(A=\dfrac{x\left(3x-1\right)}{\left(3x\right)^2-2.3x.1+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)

Thay x = - 8 và biểu thức A ta được :

\(A=\dfrac{-8}{3.\left(-8\right)-1}=\dfrac{8}{25}\)

Vậy giá trị của biểu thức A là \(\dfrac{8}{25}\) tại x = - 8

b ) Gọi \(B=\dfrac{x^2+3x+2}{x^3+2x^2-x-2}\)

Ta có \(B=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{x^2\left(x+2\right)-\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(x^2-1\right)\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x-1}\)

Thay x = 1000001 và biểu thức B ta được :

\(B=\dfrac{1}{1000001-1}=\dfrac{1}{100000}\)

Vậy giá trị của biểu thức B là \(\dfrac{1}{1000000}\) tại x = 1000001

Có 2x² + 3x + 1 = 0

<=> 2x² + 2x + x + 1 = 0

<=> 2x(x + 1) + (x + 1) = 0

<=> (2x + 1)(x + 1) = 0

=> 2x + 1 = 0 hoặc x + 1 = 0

=> x = - 1/2 hoặc x = - 1

Vậy x = - 1/2 ; x = - 1 là nghiệm của đa thức f(x)