Toán câu 1 : a ) x ² – xy + 3x – 3y b ) 3x ( x + 1 ) + x + 1 = 0 20/11/2021 By Madelyn câu 1 : a ) x ² – xy + 3x – 3y b ) 3x ( x + 1 ) + x + 1 = 0
@Peoss `a ) x ² – xy + 3x – 3y=x(x-y)+3(x-y)=(x+3)(x-y)` `b ) 3x ( x + 1 ) + x + 1 = 0` `⇔(x+1)(3x+1)=0` `+)x+1=0 ⇔x=-1` `+)3x+1=0⇔3x=-1⇔x=-1/3` Vậy `x∈{-1/3;-1}` Trả lời
Đáp án: $\underline{\text{Bạn tham khảo !!!}}$ Giải thích các bước giải: a, $x^2-xy+3x-3y$ $=x(x-y)+3(x-y)$ $=(x-y)(x+3)$ b, $3x(x+1)+x+1=0$ $\to (x+1)(3x+1)=0$ $\to \left[\begin{array}{l}x+1=0\\3x+1=0\end{array}\right.$ $\to \left[\begin{array}{l}x=-1\\x=-\dfrac{1}{3}\end{array}\right.$ Vậy $x\in \{-1;-\dfrac{1}{3} \}$ Trả lời
@Peoss
`a ) x ² – xy + 3x – 3y=x(x-y)+3(x-y)=(x+3)(x-y)`
`b ) 3x ( x + 1 ) + x + 1 = 0`
`⇔(x+1)(3x+1)=0`
`+)x+1=0 ⇔x=-1`
`+)3x+1=0⇔3x=-1⇔x=-1/3`
Vậy `x∈{-1/3;-1}`
Đáp án:
$\underline{\text{Bạn tham khảo !!!}}$
Giải thích các bước giải:
a,
$x^2-xy+3x-3y$
$=x(x-y)+3(x-y)$
$=(x-y)(x+3)$
b,
$3x(x+1)+x+1=0$
$\to (x+1)(3x+1)=0$
$\to \left[\begin{array}{l}x+1=0\\3x+1=0\end{array}\right.$
$\to \left[\begin{array}{l}x=-1\\x=-\dfrac{1}{3}\end{array}\right.$
Vậy $x\in \{-1;-\dfrac{1}{3} \}$