\(\left(x-3\right)\left(4-x\right)>0\)

\(\Rightarrow\)\(\hept{\begin{cases}x-3>0\\4-x>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>3\\x< 4\end{cases}}\)  (vô lí)

hoặc    \(\hept{\begin{cases}x-3< 0\\4-x< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< 3\\x>4\end{cases}}\)(vô lí)

Vậy      \(x=\Phi\)

a

Bài 1.

Lộn câu rồi bạn ơi..

bạn tách ra từng ít câu 1 thôi ạ

a: \(=5x\left(xy^2+3x+6y^2\right)\)

b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)

c: \(=\left(x-3\right)\left(x-4\right)\)

d: \(=x\left(x^2-2xy+y^2-9\right)\)

=x(x-y-3)(x-y+3)

e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

f: \(=\left(x-4\right)\left(x+3\right)\)

\(1,=x\left(x+y\right)+\left(x+y\right)=\left(x+1\right)\left(x+y\right)\\ 2,\Leftrightarrow x^2-3x-4x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

=(x²+x)+(xy+y)

=x(x+1)+y(x+1)

=(x+y)(x+1)

\(PT\Leftrightarrow2\left(x+7\right)-x\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-7;2\right\}\)

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)