Đặt t = x 2 – 4x ta được

t 2   +   8 t   +   15   =   t 2   +   3 t   +   5 t   +   15 = t(t + 3) + 5(t + 3) = (t + 5)(t + 3)

=   ( x 2   –   4 x   +   5 ) ( x 2   –   4 x   +   3 )   =   ( x 2   –   4 x   +   5 ) ( x 2   –   3 x   –   x   +   3 ) =   ( x 2   –   4 x   +   5 ) ( x ( x   –   3 )   –   ( x   –   3 ) )     =   ( x 2   –   4 x   +   5 ) ( x   –   1 ) ( x   –   3 )

Vậy số cần điền là -3

Đáp số cần chọn là: A

Đặt \(x^2-4x=t\)

\(\Rightarrow t^2-8t+15=0\)

\(\Leftrightarrow t^2-3t-5t+15=0\)

\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)

\(\Leftrightarrow\left(t-3\right)\left(t-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=5\\t=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4x=5\\x^2-4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=0\\x^2-4x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-5x-5=0\\x^2-4x+4-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)-5\left(x+1\right)=0\\\left(x-2\right)^2-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(x-5\right)=0\\\left(x-2-\sqrt{7}\right)\left(x-2+\sqrt{7}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\\x=2+\sqrt{7}\\x=2-\sqrt{7}\end{matrix}\right.\)

minh moi hoc lop 5 thoi

1) 3x - 6=  5x + 2

5x - 3x = -6 - 2

2x = -8

x = -4

2) 15 - x = 4x - 5

4x + x = 15 + 5

5x = 20

x = 4

Tương tự như trên 

a) -3x + 4x = 9 - 15

x = -6

Vậy x = -6

b) I3xI - 4 = 8

I3xI = 8 + 4

I3xI = 12

\(\Rightarrow\hept{\begin{cases}3x=12\\3x=-12\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=-4\end{cases}}}\)

Vậy\(x\in\left\{4;-4\right\}\)

15-3x=9-4x

15-9=4x+3x

6=7x

 x=6/7

vậy x=6/7