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\(x=\left\{x\in N;2x\left(2x-6\right)\left(3x-15\right)\right\}=0\)
\(\Leftrightarrow2x=0\Rightarrow x=0\)
\(2x-6=0\Rightarrow2x=6\Rightarrow x=3\)
\(3x-15=0\Rightarrow3x=15\Rightarrow x=5\)
\(\Rightarrow x=\left\{0;3;5\right\}\)
a) x + 123 = 234
=> x = 234 - 123
=> x = 111
b) 205 – x = 15
=> x = 205 - 15
=> x = 19
c) 2x – 10 = 60
=> 2x = 70
=> x = 35
d) x : 34 = 10
=> x = 10 * 34
=> x = 340
e) 12 + (5 + x) = 20
=> 5 + x = 8
=> x = 3
f) 130 – (100 + x) = 25
=> 100 + x = 130 - 25
=> x = 105 - 100
=> x = 5
g) 15x – 133 = 17
=> 15x = 150
=> x = 10
h) 175 + (30 – x) = 200
=> 30 – x = 25
=> x = 5
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 1: Tìm x
a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7
c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10
Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10.
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
a) x^15= x.1000^0 b ) x^15 = x.x^0
=>x^15= x . 1 =>x^15 = x.1
=> x^15 = x => x^15 = x
=> x = 1 => x = 1