\(2x\left(x-3\right)=x^2-3x\)

\(\Rightarrow2x\left(x-3\right)=x\left(x-3\right)\)

\(\Rightarrow2x=x\)

\(\Rightarrow x=0\)

\(2x.\left(x-3\right)=x^2-3x\)

\(\left(x-3\right)=x^2-3x:2x\)

Ta có: \(2x\left(x-3\right)=x^2-3x\)

\(\Leftrightarrow2x\left(x-3\right)-x\left(x-3\right)=0\)

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

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

(x+2)² +x(x-1)<2x²+1
x²+4x+4+x²-x<2x²+1
3x+4<1
x< -1

Để Q(x) có nghiệm thì Q(x) = 0

Hay: \(2x^2-3x+1=0\)

\(\Rightarrow2x^2-2x-x+1=0\)

\(\Rightarrow2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy...

`2x^2-3x+1=0`

`<=>2x^2-x-2x+1=0`

`<=>x(2x-1)-(2x-1)=0`

`<=>(2x-1)(x-1)=0`

`<=>x=1\or\x=1/2`

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

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

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

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

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

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

Áp dụng công thức là ra😎

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

B = | 2x -10| + (5-x)² + 23

|2x-10| = 2.|x-5| \(\ge\) 0 ; (5-x)²\(\ge\) 0 \(\Rightarrow\) |2x-10| + (5-x)² + 23 \(\ge\) 23

B(min) = 23 dấu bằng xảy ra khi :

\(\left\{{}\begin{matrix}|x-5|=0\\5-x=0\end{matrix}\right.\)

\(\Rightarrow\) x = 5 

kết luận B(min) = 23 xảy ra khi x = 5