35×(x-10)=32
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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
\(32.\left(10-35\right)-35.\left(18-32\right)\)
\(=32.-25-35.-14\)
\(=32.\left(-25--14\right)\)
\(=32.-11\)
\(=-352\)
c: 90\(⋮\)x-2
=>\(x-2\in\){1;-1;2;-2;3;-3;5;-5;6;-6;9;-9;10;-10;15;-15;18;-18;30;-30;45;-45;90;-90}
=>x\(\in\){3;1;4;0;5;-1;7;-3;8;-4;11;-7;12;-8;17;-13;20;-16;32;-28;47;-43;92;-88}
mà 10<=x<=90
nên \(x\in\left\{11;12;17;20;32;47\right\}\)
d: \(x⋮12\)
=>\(x\in\left\{0;12;24;36;48;60;72;...\right\}\)
mà 30<=x<=60
nên \(x\in\left\{36;48;60\right\}\)
a: x-140:35=270
=>x-4=270
=>x=270+4=274
b: \(\left(3x-1\right)^2-3^2=4^2\)
=>(3x-1)2-9=16
=>\(\left(3x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
35 . (x - 10) = 32
=> x - 10 = 32 : 35`
`=> x = 32 : 35 + 10`
`=> x =` \(\dfrac{382}{35}\)
Vậy `x=` \(\dfrac{382}{35}\)