Toán Lớp 6: a, 3 .(x+8)=17 +(-23)
b, 2.(x – 3)^2 = (-3)^2-1 ( nhiều Trường hợp)
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Toán Lớp 6: a, 3 .(x+8)=17 +(-23)
b, 2.(x – 3)^2 = (-3)^2-1 ( nhiều Trường hợp)
Comments ( 2 )
a, 3(x+8) = 17 + (-23)
⇒ 3x + 24 = -6
⇒ 3x = -6 – 24
⇒ 3x = -30
⇒ x = -30 : 3
⇒ x = -10
Vậy x = -10
b, 2(x – 3)^2 = (-3)^2 – 1
⇒ 2(x – 3)^2 = 9 – 1
⇒ 2(x – 3)^2 = 8
⇒ (x – 3)^2 = 8 : 2
⇒ (x – 3)^2 = 4
⇒ \(\left[ \begin{array}{l}x-3=2\\x-3=-2\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x = 2 +3 \\x = -2 +3\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x = 5 \\x = 1\end{array} \right.\)
Vậy x ∈ {5; 1}
Answer
a, 3 . (x + 8) = 17 + (-23)
=> 3 . (x + 8) = -6
=> (x + 8) = -6 : 3
=> x + 8 = -2
=> x = -2 – 8
=> x = -10
Vậy x = -10
________________________
b, 2 . (x – 3)^2 = (-3)^2 – 1
=> 2 . (x – 3)^2 = 9 – 1
=> 2 . (x – 3)^2 = 8
=> (x – 3)^2 = 8 : 2
=> (x – 3)^2 = 4
=> (x – 3)^2 = (+-2)^2
=> x – 3 = +-2
=> [(x – 3 = 2),(x – 3 = -2):}
=> [(x = 2 + 3),(x = -2 + 3):}
=> [(x = 5),(x = 1):}
Vậy x \in {5 ; 1}