Tìm x:
a, 11x (15 - x )=5
b, x - 43 = (57 - x) - 50
c, ( x-2 ) (8-x) = 0
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O
25 tháng 3 2022
a) \(\dfrac{9}{7}\div x=\dfrac{3}{5}\)
\(x=\dfrac{9}{7}\div\dfrac{3}{5}\)
\(x=\dfrac{15}{7}\)
b) \(x+\dfrac{1}{3}=\dfrac{8}{6}-\dfrac{1}{2}\)
\(x+\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{1}{3}\)
\(x=\dfrac{1}{2}\)
CM
5
13 tháng 7 2023
\(6\cdot x+9=15\)
\(\Rightarrow6\cdot x=15-9\)
\(\Rightarrow6\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{6}=1\)
_______________
\(43+2x=49\)
\(\Rightarrow2x=49-43\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
____________
\(x:2+5=11\)
\(\Rightarrow x:2=11-5\)
\(\Rightarrow x:2=6\)
\(\Rightarrow x=6\cdot2=12\)
______________
\(77-11x=0\)
\(\Rightarrow11x=77\)
\(\Rightarrow x=\dfrac{77}{11}\)
\(\Rightarrow x=7\)
_______________
\(12-4:x=8\)
\(\Rightarrow4:x=12-8\)
\(\Rightarrow4:x=4\)
\(\Rightarrow x=\dfrac{4}{4}=1\)
_____________
\(x:3+8=11\)
\(\Rightarrow x:3=11-8\)
\(\Rightarrow x:3=3\)
\(\Rightarrow x=3\cdot3=9\)
13 tháng 7 2023
a: 6x+9=15
=>6x=6
=>x=1
b: 2x+43=49
=>2x=6
=>x=3
c: x:2+5=11
=>x:2=6
=>x=12
d: 77-11x=0
=>7-x=0
=>x=7
e: 12-4:x=8
=>4:x=4
=>x=1
f: x:3+8=11
=>x:3=3
=>x=9
9 tháng 7 2023
`@` `\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}`
8 tháng 7 2023
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) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
ND
Nguyễn Đức Trí
VIP
8 tháng 7 2023
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
12 tháng 8 2021
Bài 2:
a: Ta có: \(2x+79=x+45\)
nên 2x-x=45-79
hay x=-24
b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)
\(\Leftrightarrow8-x-x-14=0\)
\(\Leftrightarrow2x=22\)
hay x=11
b) Ta có: \(x-43=\left(57-x\right)-50\)
\(\Leftrightarrow x-43=57-x-50\)
\(\Leftrightarrow x+x=7+43\)
\(\Leftrightarrow2x=50\)
hay x=25
Vậy: x=25
c) Ta có: (x-2)(8-x)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\8-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
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