a) Ta có M = (x + 3)2 - (x - 1)(x + 4) + 5
= x2 + 6x + 9 -(x2 + 3x - 4) + 5
= x2 + 6x + 9 - x2 - 3x + 4 + 5
= 3x + 18 (1)
b) Thay x = 2 vào (1)
=> M = 3.2 + 18 = 24
c) Ta có M = 15x2
=> 15x2 = 3x + 18
=> 15x2 - 3x - 18 = 0
=> 15x2 + 15x - 18x - 18 = 0
=> 15x(x + 1) - 18(x + 1) = 0
=> (15x - 18)(x + 1) = 0
=> 3(5x - 6)(x + 1) = 0
=> (5x - 6)(x + 1) = 0
=> \(\orbr{\begin{cases}5x-6=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1,2\\x=-1\end{cases}}\)
Vậy \(x\in\left\{1,2;-1\right\}\)là giá trị cần tìm
a, \(M=\left(x+3\right)^2-\left(x-1\right)\cdot\left(x+4\right)+5\)\(=x^2+6x+9-\left(x^2-x+4x-4\right)+5\)\(=3x+18\)
b, Thay x=2 vào M có \(M=3\cdot2+18=24\)
c, \(M=15x^2\Leftrightarrow15x^2=3x+18\Leftrightarrow15x^2-3x-18=0\Leftrightarrow3\cdot\left(x+1\right)\cdot\left(5x-6\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x=-1\\x=\frac{6}{5}\end{cases}}\)
Vậy ....
M = ( x + 3 )2 - ( x - 1 )( x + 4 ) + 5
= x2 + 6x + 9 - ( x2 + 3x - 4 ) + 5
= x2 + 6x + 14 - x2 - 3x + 4
= 3x + 18
* Khi x = 2 => M= 3.2 + 18 = 6 + 18 = 24
* Để M = 15x2
=> 15x2 = 3x + 18
<=> 15x2 - 3x - 18 = 0
<=> 3( 5x2 - x - 6 ) = 0
<=> 5x2 - x - 6 = 0
<=> 5x2 + 5x - 6x - 6 = 0
<=> 5x( x + 1 ) - 6( x + 1 ) = 0
<=> ( x + 1 )( 5x - 6 ) = 0
<=> x + 1 = 0 hoặc 5x - 6 = 0
<=> x = -1 hoặc x = 6/5
Vậy S = { -1 ; 6/5 }