a, \(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\end{matrix}\right.\)

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

Vậy ...

b, \(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy ...

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

\(\Leftrightarrow\left(x+2\right)\left(x+2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\)

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

Vậy ...

\(a.\)

\(3x^2-6x=0\)

\(\Leftrightarrow3x\cdot\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\end{matrix}\right.\)

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

\(b.\)

\(x\cdot\left(x-6\right)+10\cdot\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\cdot\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

\(c.\)

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

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

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

\(\Leftrightarrow\left(x+1\right)\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)

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

\(Q=2x^2-6x\)

\(=2\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)\)

\(=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge\frac{-9}{2}\forall x\)

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

\(M=x^2+y^2-x+6x+10\)

\(=x^2+y^2+5x+10\)

\(=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+10+y^2\)

\(=\left(x+\frac{5}{2}\right)^2+y^2+\frac{15}{4}\)

Vì \(\hept{\begin{cases}\left(x+\frac{5}{2}\right)^2\ge0;\forall x,y\\y^2\ge0;\forall x,y\end{cases}}\)

\(\Rightarrow\left(x+\frac{5}{2}\right)^2+y^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+\frac{5}{2}\right)^2+y^2+\frac{15}{4}\ge0+\frac{15}{4};\forall x,y\)

Hay \(M\ge\frac{15}{4};\forall x,y\)

Dấu =" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{5}{2}\right)^2=0\\y^2=0\end{cases}}\)

                      \(\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{2}\\y=0\end{cases}}\)

Vậy MIN \(M=\frac{15}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{2}\\y=0\end{cases}}\)

\(x^2-6x+10x=x^2-3x-3x+9+1=x\left(x-3\right)-3\left(x-3\right)+1=\left(x-3\right)\left(x-3\right)+1=\left(x-3\right)^2+1\)

mà \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1\)

Vậy không tìm được x thỏa mãn yêu cầu đề.

Dễ mà :vv

Ta có: \(x^2+4y^2-6x+4y+10=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(2y-1\right)^2=0\)

Đến đây tự giải...

<=> x^2-6x+9+4y^2+4y+1=0

<=> x^2-2.3.x+3^2+(2y)^2+2.2y.1+1=0

<=>(x-3)^2+(2y+1)^2=0

<=> x-3=0 và 2y+1=0

<=> x=3 và y=-1/2

\(\text{A.}\)\(\text{x3+6x2+3x−10}\)

5x - (x + 2) = -6x + 10

=> 5x - x - 2 = -6x + 10

=> 4x + 6x = 10 + 2

=> 10x = 12

=> x = 12:10

=> x = 1,2

X = 4 PHẢI KO BẠN NHỈ 

NẾU ĐÚNG CHO MK LI-KE NHA !

X = 4

ĐÚNG KO BẠN

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