Tìm x: a, x.(x+7)=0 b, (x+12).(x-3)=0 c, (-x+5).(3-x)=0 d, x.(2+x).(7-x)=0 e,(x-1).(x+2).(-x-3)=0 25/10/2021 Bởi Cora Tìm x: a, x.(x+7)=0 b, (x+12).(x-3)=0 c, (-x+5).(3-x)=0 d, x.(2+x).(7-x)=0 e,(x-1).(x+2).(-x-3)=0
`a//x.(x+7)=0` `⇒` \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\) Vậy `x∈{0;-7}` `b//(x+12).(x-3)=0` `⇒` \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\) Vậy `x∈{-12;3}` `c//(-x+5).(3-x)=0` `⇒` \(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}-x=-5\\x=3\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\) Vậy `x∈{5;3}` `d//x.(2+x).(7-x)=0` `⇒` \(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\) Vậy `x∈{0;-2;7}` `e//(x-1).(x+2).(-x-3)=0` `⇒` \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=1\\x=-2\\-x=3\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\) Vậy `x∈{1;-2;-3}` Chúc bạn học tốt nha !!! @MiCi2612 Bình luận
a) x . ( x + 7 ) = 0 <=> \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\) Vậy x = 0 hoặc x = -7 . b) ( x + 12 ) . ( x – 3 ) = 0 <=> \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\) Vậy x = -12 hoặc x = 3. c) ( -x + 5 ) . ( 3 – x ) = 0 <=> \(\left[ \begin{array}{l}-x+5=0\\3-x = 0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\) Vậy x = 5 hoặc x = 3. d) x . ( 2 + x ) . ( 7 – x ) = 0 <=>\(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\) Vậy x = 0 hoặc x = -2 hoặc x = 7 e) ( x – 1 ) ( x + 2 ) ( -x – 3 ) = 0 <=> \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3 =0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\) Vậy x = 1 hoặc x = -2 hoặc x = -3. Bình luận
`a//x.(x+7)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy `x∈{0;-7}`
`b//(x+12).(x-3)=0`
`⇒` \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy `x∈{-12;3}`
`c//(-x+5).(3-x)=0`
`⇒` \(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}-x=-5\\x=3\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy `x∈{5;3}`
`d//x.(2+x).(7-x)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy `x∈{0;-2;7}`
`e//(x-1).(x+2).(-x-3)=0`
`⇒` \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=1\\x=-2\\-x=3\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
Vậy `x∈{1;-2;-3}`
Chúc bạn học tốt nha !!!
@MiCi2612
a) x . ( x + 7 ) = 0
<=> \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy x = 0 hoặc x = -7 .
b) ( x + 12 ) . ( x – 3 ) = 0
<=> \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy x = -12 hoặc x = 3.
c) ( -x + 5 ) . ( 3 – x ) = 0
<=> \(\left[ \begin{array}{l}-x+5=0\\3-x = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy x = 5 hoặc x = 3.
d) x . ( 2 + x ) . ( 7 – x ) = 0
<=>\(\left[ \begin{array}{l}x=0\\2+x=0\\7-x=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy x = 0 hoặc x = -2 hoặc x = 7
e) ( x – 1 ) ( x + 2 ) ( -x – 3 ) = 0
<=> \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3 =0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
Vậy x = 1 hoặc x = -2 hoặc x = -3.