bai?: tim x biet a) 2.x – 15 = -21 b)12 – (30-x) =-23 c)x.(x -3) = 0 d) ║x + 7 ║ – 2 = 0 20/07/2021 Bởi Amara bai?: tim x biet a) 2.x – 15 = -21 b)12 – (30-x) =-23 c)x.(x -3) = 0 d) ║x + 7 ║ – 2 = 0
Đáp án: Giải thích các bước giải: a)2x-15=-21 b)12-(30-x)=-23 c)x.(x-3)=0 d)║x+7║-2=0 ⇔2x =-21+15 ⇔30+x =-23-12 ⇔x²-3x=0 th1:x+7-2=0 ⇔2x =-6 ⇔30+x =-35 ⇔x² ⇔x+5=0 ⇔x =-3 ⇔x =-65 (câu này hong bt^^) ⇔x=-5 th2:x+7+2=0 ⇔x+9=0 ⇔x=-9 câu c) mik hok bt xin lỗi bạn nhas^^! Bình luận
Bạn tham khảo nhé a) 2.x – 15 = -21 <=> 2.x = -21 + 15 <=> 2.x = -6 <=> x = -3 Vậy x = -3 b) 12 – (30 – x) = -23 <=> 12 – 30 + x = -23 <=> -18 + x = -23 <=> x = -23 + 18 <=> x = -5 Vậy x = -5 c) x.(x – 3) = 0 <=> \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) Vậy \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) d) ║x + 7║ – 2 = 0 <=> ║x + 7║ = 2 (1) ║x + 7║ = $\left \{ {{x+7 khix+7\geq 0 <=> x\geq-7} \atop {-x-7 khi x+7<0<=> x<-7}} \right.$ (1) <=> \(\left[ \begin{array}{l}\left \{ {{x\geq-7} \atop {x+7=2}} \right.\\\left \{ {{x<-7} \atop {-x-7=2}} \right.\end{array} \right.\) <=> \(\left[ \begin{array}{l}\left \{ {{x\geq-7} \atop {x=-5(TM) }} \right.\\\left \{ {{x<-7} \atop {x=-9(TM)}} \right.\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=-5\\x=-9\end{array} \right.\) Vậy \(\left[ \begin{array}{l}x=-5\\x=-9\end{array} \right.\) Bình luận
Đáp án:
Giải thích các bước giải:
a)2x-15=-21 b)12-(30-x)=-23 c)x.(x-3)=0 d)║x+7║-2=0
⇔2x =-21+15 ⇔30+x =-23-12 ⇔x²-3x=0 th1:x+7-2=0
⇔2x =-6 ⇔30+x =-35 ⇔x² ⇔x+5=0
⇔x =-3 ⇔x =-65 (câu này hong bt^^) ⇔x=-5
th2:x+7+2=0
⇔x+9=0
⇔x=-9
câu c) mik hok bt xin lỗi bạn nhas^^!
Bạn tham khảo nhé
a) 2.x – 15 = -21
<=> 2.x = -21 + 15
<=> 2.x = -6
<=> x = -3
Vậy x = -3
b) 12 – (30 – x) = -23
<=> 12 – 30 + x = -23
<=> -18 + x = -23
<=> x = -23 + 18
<=> x = -5
Vậy x = -5
c) x.(x – 3) = 0
<=> \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
d) ║x + 7║ – 2 = 0
<=> ║x + 7║ = 2 (1)
║x + 7║ = $\left \{ {{x+7 khix+7\geq 0 <=> x\geq-7} \atop {-x-7 khi x+7<0<=> x<-7}} \right.$
(1) <=> \(\left[ \begin{array}{l}\left \{ {{x\geq-7} \atop {x+7=2}} \right.\\\left \{ {{x<-7} \atop {-x-7=2}} \right.\end{array} \right.\)
<=> \(\left[ \begin{array}{l}\left \{ {{x\geq-7} \atop {x=-5(TM) }} \right.\\\left \{ {{x<-7} \atop {x=-9(TM)}} \right.\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-5\\x=-9\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=-5\\x=-9\end{array} \right.\)