Toán (100.x-10).(6+x)=0 (6.x+12).(15-5.x)=0 x.(x+7)-15.(x+7)=0 01/07/2021 By Reagan (100.x-10).(6+x)=0 (6.x+12).(15-5.x)=0 x.(x+7)-15.(x+7)=0
(100.x-10).(6+x)=0 ⇒ \(\left[ \begin{array}{l}100x-10=0 ⇔ 100x =10 ⇔ x = \frac{10}{100} = 0,1 \\6+x=0 ⇔ x=-6\end{array} \right.\) (6.x+12).(15-5.x)=0 ⇒ \(\left[ \begin{array}{l}6×x +12=0 ⇔ 6×x =-12 ⇔ x = -2 \\15-5×x=0 ⇔ 5×x = 15 ⇔ x = 3 \end{array} \right.\) x.(x+7)-15.(x+7)=0 = (x-15) . (x+7) = 0 ⇒ \(\left[ \begin{array}{l}x-15=0 ⇔ x =15 \\x+7=0 ⇔ x = -7 \end{array} \right.\) Trả lời
Đáp án: +) `(100.x-10).(6+x)=0` `⇒` 2 Trường hợp: TH1: `100 . x – 10 = 0` `100 . x = 0 + 10` `100 . x = 10` `x = 10 : 100` `x = 0,1` TH2: `6 + x = 0` `x = 0 – 6` `x = -6` Vậy `x ∈ {0,1 ; -6}` +) `(6.x+12).(15-5.x)=0` `⇒` 2 Trường hợp: TH1: `6x + 12 = 0` `6x = 0 – 12` `6x = -12` `x = -12 : 6` `x = -2` TH2: 15 – 5 . x = 0` `5x = 15 – 0` `5x = 15` `x = 15 : 5` `x = 3` Vậy `x ∈ { – 2;3}` +) `x.(x+7)-15.(x+7)=0` `(x – 15) . (x + 7) = 0` `⇒ 2` Trường hợp: TH1: `x – 15 = 0` `x = 0 +15` `x = 15` TH2: `x + 7 = 0` `x = 0 – 7` `x = – 7` Vậy `x ∈ {15 ; -7}` Trả lời
(100.x-10).(6+x)=0
⇒ \(\left[ \begin{array}{l}100x-10=0 ⇔ 100x =10 ⇔ x = \frac{10}{100} = 0,1 \\6+x=0 ⇔ x=-6\end{array} \right.\)
(6.x+12).(15-5.x)=0
⇒ \(\left[ \begin{array}{l}6×x +12=0 ⇔ 6×x =-12 ⇔ x = -2 \\15-5×x=0 ⇔ 5×x = 15 ⇔ x = 3 \end{array} \right.\)
x.(x+7)-15.(x+7)=0
= (x-15) . (x+7) = 0
⇒ \(\left[ \begin{array}{l}x-15=0 ⇔ x =15 \\x+7=0 ⇔ x = -7 \end{array} \right.\)
Đáp án:
+) `(100.x-10).(6+x)=0`
`⇒` 2 Trường hợp:
TH1: `100 . x – 10 = 0`
`100 . x = 0 + 10`
`100 . x = 10`
`x = 10 : 100`
`x = 0,1`
TH2: `6 + x = 0`
`x = 0 – 6`
`x = -6`
Vậy `x ∈ {0,1 ; -6}`
+) `(6.x+12).(15-5.x)=0`
`⇒` 2 Trường hợp:
TH1: `6x + 12 = 0`
`6x = 0 – 12`
`6x = -12`
`x = -12 : 6`
`x = -2`
TH2: 15 – 5 . x = 0`
`5x = 15 – 0`
`5x = 15`
`x = 15 : 5`
`x = 3`
Vậy `x ∈ { – 2;3}`
+) `x.(x+7)-15.(x+7)=0`
`(x – 15) . (x + 7) = 0`
`⇒ 2` Trường hợp:
TH1: `x – 15 = 0`
`x = 0 +15`
`x = 15`
TH2: `x + 7 = 0`
`x = 0 – 7`
`x = – 7`
Vậy `x ∈ {15 ; -7}`