Toán Lớp 8: Tìm x
$1) (x+2)^2+3(x-5)-x^2=3$
$2) x(x+6)+7x+42=0$
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Toán Lớp 8: Tìm x
$1) (x+2)^2+3(x-5)-x^2=3$
$2) x(x+6)+7x+42=0$
Comments ( 2 )
Giải đáp + Lời giải và giải thích chi tiết:
1)
(x+2)^2+3(x-5)-x^2=3
⇔ (x^2+4x+4)+(3x-15)-x^2-3=0
⇔ x^2+4x+4+3x-15-x^2-3=0
⇔ (x^2-x^2)+(4x+3x)+(4-15-3)=0
⇔ 7x-14=0
⇔ 7x=14
⇔ x=2
Vậy x=2
b)
x(x+6)+7x+42=0
⇔ x(x+6)+(7x+42)=0
⇔ x(x+6)+7(x+6)=0
⇔ (x+6)(x+7)=0
⇔ $\left[\begin{matrix} x+6=0\\ x+7=0\end{matrix}\right.$
⇔ $\left[\begin{matrix} x=-6\\ x=-7\end{matrix}\right.$
Vậy x=-6 hoặc x=-7
$#Hy$
1) (x + 2)^2 + 3(x-5) – x^2 = 3
<=> x^2 + 2×2 + 2^2 + 3x – 3.5 – x^2 – 3= 0
<=> x^2 + 4x + 4 + 3x – 15 – x^2 – 3 = 0
<=> 7x – 14 = 0
<=> 7x = 14
=> x = 2
2) x(x+6) + 7x + 42 = 0
<=> x(x+6) + (7x + 42) = 0
<=> x(x+6) + 7(x + 6) = 0
<=> (x+7)(x+ 6) = 0
<=> \(\left[ \begin{array}{l}x+7 = 0\\x+6=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-7\\x=-6\end{array} \right.\)