Để phương trình vô nghiệm thì \(m\in\varnothing\)

6m 3 cm < 7m

6m 3 cm > 6m

6m 3 cm < 630cm

6m 3 cm = 603cm

5 m 6 cm > 5m

5 m 6 cm < 6m

5 m 6 cm = 506cm

5 m 6 cm < 560cm.

\(y=\dfrac{2x-1}{3x-6m}\Rightarrowđkxđ:\Leftrightarrow3x-6m\ne0\Leftrightarrow x\ne2m\)

\(xét\)\(\) \(x\) \(trên\) \(khoảng\) \(\left(0;1\right)\)\(\Rightarrow x\ne2m\Leftrightarrow\left[{}\begin{matrix}2m\le0\\2m\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\le0\\m\ge\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow m\in(-\text{∞};0]\cup[\dfrac{1}{2}:+\text{∞})\)

a: Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}\dfrac{1}{3}x+m+\dfrac{1}{3}=2x-6m+5\\y=\dfrac{1}{3}x+m+\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{3}x=-7m+5\\y=\dfrac{1}{3}x+m+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21}{5}m-3\\y=\dfrac{1}{3}\left(\dfrac{21}{5}m-3\right)+m+\dfrac{1}{3}=\dfrac{7}{5}m-1+m+\dfrac{1}{3}=\dfrac{12}{5}m-\dfrac{2}{3}\end{matrix}\right.\)

 b: Theo đề, ta có: \(\dfrac{12}{5}m-\dfrac{2}{3}=9\cdot\left(\dfrac{21}{5}m-3\right)^2\)

Đến đây bạn chỉ cần giải phương trình bậc hai ra thôi

a, \(m^2-6m+x^2-x+3\)

\(=m^2-3m-3m+9+x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{25}{4}\)

\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\)

Với mọi giá trị của \(m;x\in R\) ta có:

\(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\)

Để \(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}=-\dfrac{25}{4}\) thì

\(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy..............

b, \(3x^2-6x+12\)

\(=3x^2-3x-3x+3+9\)

\(=3x\left(x-1\right)-3\left(x-1\right)+9\)

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

Với mọi giá trị của \(x\in R\) ta có:

\(3\left(x-1\right)^2+9\ge9\)

Để \(3\left(x-1\right)^2+9=9\) thì

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

Vậy..............

Chúc bạn học tốt!!!

a, \(A=m^2-6m+x^2-x+3\)

\(=x^2-6m+9+x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{25}{4}\)

\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(MIN_A=\dfrac{-25}{4}\) khi m = 3, \(x=\dfrac{1}{2}\)

b, \(B=3x^2-6x+12=3\left(x^2-2x+4\right)\)

\(=3\left(x^2-2x+1+3\right)=3\left(x-1\right)^2+9\ge9\)

Dấu " = " khi \(3\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy MIN B = 9 khi x = 1

Bài 1: 

a:9x-6=3x+12

=>6x=18

hay x=3

b: \(\Leftrightarrow5\left(x+1\right)-8x=50\)

=>5x+5-8x=50

=>-3x+5=50

=>-3x=45

hay x=-15

c: \(\Leftrightarrow7\left(x+1\right)+2x^2=x+23\)

\(\Leftrightarrow2x^2+7x+7-x-23=0\)

\(\Leftrightarrow2x^2+6x-16=0\)

\(\Leftrightarrow x^2+3x-8=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-8\right)=9+32=41>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{41}}{2}\\x_2=\dfrac{-3+\sqrt{41}}{2}\end{matrix}\right.\)