a) đk: \(x\ne\left\{0;2\right\}\)

Ta có:

\(M=\frac{x}{x-2}\div\frac{2x}{x^2-2x}\)

\(M=\frac{x}{x-2}\cdot\frac{x\left(x-2\right)}{2x}\)

\(M=\frac{x}{2}\)

b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=3\end{cases}}\)

Tại x = 3 thì giá trị của M là: \(M=\frac{3}{2}\)

c) Để \(M\ge0\Leftrightarrow\frac{x}{2}\ge0\Rightarrow x\ge0\)

Vậy khi \(x\ge0\Leftrightarrow M\ge0\)

Tính giá trị biểu thức 2x^2-3x+1. Tại x thõa mãn x^2=1/4

1, Thay x = 1/3 ; y = -1/5 ta được 

\(=\dfrac{3.1}{9}-5\left(-\dfrac{1}{5}\right)+1=\dfrac{1}{3}+2=\dfrac{7}{3}\)

2, Thay x = -2 ; y = -1/2 ta được 

\(=5.4\left(-\dfrac{1}{2}\right)+3\left(-2\right)\left(-\dfrac{1}{2}\right)-\dfrac{2\left(-2\right).1}{4}\)

\(=-10+3+1=-6\)

cảm ơn xong chẳng có ai :)))

1./ \(x+y=3\Rightarrow\left(x+y\right)^3=27\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\Rightarrow x^3+y^3+3\cdot2\cdot3=27.\)

\(\Rightarrow x^3+y^3=9\)

2./ \(\left(x+3\right)\left(x^2-3x+3^2\right)-x^3-2x-4=0\)

\(\Leftrightarrow x^3+27-x^3-2x-4=0\Leftrightarrow2x=23\Leftrightarrow x=\frac{23}{2}\)

1/ \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow x^2+4+y^2=9\)

\(\Rightarrow x^2+y^2=5\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.1=3\)

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

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

Với \(x\ne-1\), ta có:

\(3x-1=0\Rightarrow3x=1\) \(\Rightarrow x=\dfrac{1}{3}\left(tm\right)\)

Thay \(x=\dfrac{1}{3}\) vào (1), ta được:

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

\(=-\dfrac{8}{3}:\dfrac{4}{3}=-\dfrac{8}{3}\cdot\dfrac{3}{4}=-2\)

Vậy: ...