Tìm x (4-2x).(x+3)=0 -x.(8-2x)=0 (2x-6).(x-3)=0 07/11/2021 Bởi Ariana Tìm x (4-2x).(x+3)=0 -x.(8-2x)=0 (2x-6).(x-3)=0
Đáp án: Giải thích các bước giải: a, ( 4 – 2x ) . ( x + 3 ) = 0 Để ( 4 – 2x ) . ( x + 3 ) = 0 <=> 2( 2 – x ) . ( x + 3 ) = 0 TH1: 2( 2 – x ) = 0 2 – x = 0 : 2 2 – x = 0 => x = 2 ( TM ) TH2: x + 3 = 0 => x = – 3 TH3: – 2x = 0 => x = 0 TH4: 4 – x = 0 => x = 4 Ta có: ( 2x – 6 ) . ( x – 3 ) = 0 <=> 2( x – 3 ) . ( x – 3 ) = 0 <=> 2( x – 3 )² = 0 => ( x – 3 )² = 0 => x – 3 = 0 =>x = 3 b, – x . ( 8 – 2x ) = 0 8 – 2x = 0 : ( – x ) 8 – 2x = 0 2x = 8 – 0 2x = 8 x = 8 : 2 x = 4 Bình luận
`(4-2x).(x+3)=0` `⇔2(2-x)(x+3)=0` `th1` `2-x=0` `⇒x=2` `th2` `x+3=0` `⇒x=-3` `-x.(8-2x)=0` `⇔-2x(4-x)=0` `th1` `-2x=0` `⇒x=0` `th2` `4-x=0` `⇒x=4` `(2x-6).(x-3)=0` `⇔2(x-3)(x-3)=0` `⇔2(x-3)^2=0` `⇒x-3=0` `⇒x=3` Bình luận
Đáp án:
Giải thích các bước giải:
a, ( 4 – 2x ) . ( x + 3 ) = 0
Để ( 4 – 2x ) . ( x + 3 ) = 0
<=> 2( 2 – x ) . ( x + 3 ) = 0
TH1: 2( 2 – x ) = 0
2 – x = 0 : 2
2 – x = 0
=> x = 2 ( TM )
TH2: x + 3 = 0
=> x = – 3
TH3: – 2x = 0
=> x = 0
TH4: 4 – x = 0
=> x = 4
Ta có: ( 2x – 6 ) . ( x – 3 ) = 0
<=> 2( x – 3 ) . ( x – 3 ) = 0
<=> 2( x – 3 )² = 0
=> ( x – 3 )² = 0
=> x – 3 = 0
=>x = 3
b, – x . ( 8 – 2x ) = 0
8 – 2x = 0 : ( – x )
8 – 2x = 0
2x = 8 – 0
2x = 8
x = 8 : 2
x = 4
`(4-2x).(x+3)=0`
`⇔2(2-x)(x+3)=0`
`th1`
`2-x=0`
`⇒x=2`
`th2`
`x+3=0`
`⇒x=-3`
`-x.(8-2x)=0`
`⇔-2x(4-x)=0`
`th1`
`-2x=0`
`⇒x=0`
`th2`
`4-x=0`
`⇒x=4`
`(2x-6).(x-3)=0`
`⇔2(x-3)(x-3)=0`
`⇔2(x-3)^2=0`
`⇒x-3=0`
`⇒x=3`