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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`
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) ( 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
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
a: \(\left(x-15\right)\cdot15=0\)
=>x-15=0
=>x=0+15
=>x=15
b: 32(x-10)=32
=>x-10=32/32=1
=>x=1+10=11
c: (x-5)(x-7)=0
=>\(\left[\begin{array}{l}x-5=0\\ x-7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=7\end{array}\right.\)
d: \(\left(x-35\right)\cdot35=35\)
=>\(x-35=\frac{35}{35}=1\)
=>x=1+35
=>x=36
575- [6x+70] = 445 315 +[125-x] = 435 f 541+[218-x]=435
6x+70 =575-445 125-x = 435- 315 218-x = 541-435
6x+70 = 130 125-x =120 218-x =106
6x = 130-70 x = 125 - 120 x = 218-106
6x = 60 x = 5 x = 112
x = 60:6
x = 10
bạn ơi khó hiểu quá,.,.,.
\(\orbr{\begin{cases}\\\end{cases}}\)
b, -32 . ( x - 5 ) = 0
<-> x - 5 = 0
<-> x = 0 + 5
<-> x = 5
Vậy x = 5
c, 11 + ( 15 - x ) = 1
<-> 15 - x = 1 - 11
<-> 15 - x = -10
<-> x = 15 - ( - 10 )
<-> x = 25
Vậy x = 25