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Bài 6. Tìm x ϵ N biết
a) (x –15) .15 = 0
b) 32 (x –10 ) = 32
c) ( x – 5)(x – 7) = 0
d) (x – 35).35 = 35
A.\(\left(x-15\right).15=0\)
\(x-15=0:15\)
\(x-15=0\)
\(x=15+0\)
\(x=15\)
B.\(32\left(x-10\right)=32\)
\(x-10=32:32\)
\(x-10=1\)
\(x=10+1\)
\(x=11\)
`a) `
`(x-15)xx15=0`
`<=> x-15 = 0 : 15`
`<=> x-15 = 0`
`<=> x = 0 + 15`
`<=> x =15`
`b)`
`32.(x-10)=32`
`<=> x - 10 = 32:32`
`<=>x-10=1`
`<=> x = 1+10`
`<=> x =11`
`c)`
`(x-5).(x-7)=0`
`<=>` \(\left[ \begin{array}{l}x-5 = 0\\x-7=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
`d)`
`(x-35)xx35=35`
`<=> x - 35 = 35:35`
`<=> x - 35 = 1`
`<=> x = 1+35`
`<=> x = 36`
a) ( x - 15 ) . 35 = 0
x - 15 = 0 : 35
x - 15 = 0
x = 0 + 15
x = 15
b) 32 ( x - 10 ) = 32
x - 10 = 32 : 32
x - 10 = 1
x = 1 + 10
x = 11
a) (x - 15) . 35 = 0
(x - 15)=0 : 35
(x - 15)=0
x=0+15
x=15
b) 32.(x - 10) = 32
(x - 10) = 32 : 32
(x - 10) = 1
x = 1 + 10
x = 11
a) (5x - 12 ) : 4 = 318 : 315
( 5x - 12 ) : 4 = 33
( 5x - 12 ) : 4 = 27
5x - 12 = 108
5x = 120
x = 24
b) 5x3 - 102 = 35
5x3 - 100 = 35
5x3 = 135
x3 = 27
x3 = 33
=> x = 3
c) 65 - 4x+2 = 20180
65 - 4x+2 = 1
4x+2 = 64
4x+2 = 43
x + 2 = 3
=> x = 1
d) 2x+1 - 2x = 32
2x . ( 2 - 1 ) = 32
2x . 1 = 32
2x = 32
2x = 25
=> x = 5
a/ (3x -12 ):8+10=13
<=>(3x-12):8=3
<=>3x-12=24
<=>3x=36
<=>x=12
b/(x+5)x8-20=60
<=>(x+5)x8=80
<=>x+5=80:8
<=>x+5=10
x=5
c/3x[35+(x-10)]-375=0
<=>3x[35+(x-10)]=375
<=>35+(x+10)=125
<=>x+10=90
x=80
d/(158x+173-24):35=81
<=>158x+173-24=2835
<=>158x+173=2859
<=>158x=2686
<=>x=17
e/x-140:35=0
<=>x-140=0
<=>x=140+0
x=140
Bài 6:
a)(x-15).15=0
⇔x-15=0
⇔x=15
b)32(x-10)=32
⇔x-10=1
⇔x=11
c)(x-5)(x-7)=0
⇔x-5=0 hay x-7=0
⇔x=5 hay x=7
d)(x-35)35=35
⇔x-35=1
⇔x=36