d) /x/=/-4/
h) 3/2x-1/=15
k) (x+12).(x.3)=0
i) /x/<5
m) 3.x mũ 3 -12+x mũ 3 =4+2.x mũ 3
d) /x/=/-4/ h) 3/2x-1/=15 k) (x+12).(x.3)=0 i) /x/<5 m) 3.x mũ 3 -12+x mũ 3 =4+2.x mũ 3
By Arya
By Arya
d) /x/=/-4/
h) 3/2x-1/=15
k) (x+12).(x.3)=0
i) /x/<5
m) 3.x mũ 3 -12+x mũ 3 =4+2.x mũ 3
Đáp án:
`d)`
l `x`l `= 4`
` =>x = 4` hoặc ` x = -4`
` h)`
`3` l`2x-1`l `= 15`
` => l`2x-1`l `= 5`
`=>` TH1
` 2X -1 = 5`
=> 2X = 6`
` => X= 3`
TH2
` 2x -1 = -5`
` => 2x = -4`
` => x = -2`
` k)`
` (x+12).(x+3) = 0`
` => x +12 = 0` hoặc ` x+ 3 = 0`
` => x =-12` hoặc ` x = -3`
` i`
l`x`l` <5`
` => x ∈ { -4;-3;-2;-1;0;1;2;3;4}`
`d, |x|=|-4|`
`|x|=4`
`=>`\(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy `x=4` hoặc `x=-4`
`h, 3|2x-1|=15`
`|2x-1|=5`
`<=>`\(\left[ \begin{array}{l}2x-1=5\\2x-1=-5\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=6\\2x=-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `x=3` hoặc `x=-2`
`k, (x+12).(x.3)=0`
`<=>`\(\left[ \begin{array}{l}x+12=0\\x.3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-12\\x=0\end{array} \right.\)
Vậy `x=-12` hoặc `x=0`
`i, |x|<5`
`+, |x|=0`
`=> x=0`
`+, |x|=1`
`=>`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
`+, |x|=2`
`=>`\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
`+,|x|=3`
`=>`\(\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\)
`+,|x|=4`
`=>`\(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy `x ∈`{`-4;-3;-2;-1;0;1;2;3;4`}
`m,3.x³ -12+x³=4+2.x³`
`3.x³+x³-2x³=-12+4`
`x³.(2-1-2)=-12+4`
`x³(-1)=-12+4`
`x³=-8`
`x³=2³`
`=>x=2`