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`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
b) \(697:\frac{15x+364}{x}=17\)
\(\frac{15x+364}{x}=697:17\)
\(\frac{15x+364}{x}=41\)
\(15+\frac{364}{x}=41\)
\(\frac{364}{x}=41-15\)
\(\frac{364}{x}=26\)
\(\Rightarrow x=14\)
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
hung pham tien bn vào câu hỏi tương tự đi , có đấy
**** cho minh
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) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
\(\frac{30x+40}{x}=1500:30\)
\(\frac{30x+40}{x}=50\)
Suy ra 30x+40= 50.x
40= 50x- 30x
40= 20x
x=40:20=2
\(1500:\frac{30x+40}{x}=30\)
\(\frac{30x+40}{x}=1500:30\)
\(\frac{30x+40}{x}=50\)
\(\Rightarrow30x+40=50x\)
\(30x+40-50x=0\)
\(40+\left(30x-50x\right)=0\)
\(40-20x=0\)
\(20\left(20-x\right)=0\)
\(\Rightarrow20-x=0\)
\(x=20\)
Vậy x = 20
1500: [(30x + 40) : x]=30
=>(30x + 40): x = 1500 : 30
=>30x : x+ 40:x=50
=> 30 + 40 :x=50
=> 40 :x=50-30
=> 40:x=20
=>x=40:20
=>x=2
: [(30x + 40) : x] = 30(2x - 15)5 = (2x - 15)3
=>(30x+40):x=1500 : 30=50
40:x = 50-30
40:x=20
x=2