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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}.`
`@` `\text {Kaizuu lv uuu}`
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
`@` `\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}`
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
a) \(4x+4=16\)
\(4x=12\)
\(x=3\)
b) \(34\left(2x-6\right)=0\)
\(2x=6\)
\(x=3\)
c) \(15:x=5\)
\(x=15:5=3\)
d) \(20-\left(x+14\right)=5\)
\(x+14=20-5=15\)
\(x=15-14=1\)
a) \(4x+4=16\)
\(\Rightarrow4x=16-4\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=\dfrac{12}{4}\)
\(\Rightarrow x=3\)
b) \(34\cdot\left(2x-6\right)=0\)
\(\Rightarrow2x-6=\dfrac{0}{36}\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}\)
\(\Rightarrow x=3\)
c) \(15:x=5\)
\(\Rightarrow x=15:5\)
\(\Rightarrow x=3\)
d) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
+Phần a:
\(\left(2x-6\right).x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy giá trị của x là : 0 hoặc 3
+Phần b:
\(\left(x+12\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-12\\x=1\end{cases}}\)
Vậy giá trị của x là : -12 hoặc 1
Phần c bạn tự làm nhé.
HỌC TỐT :))
Ta có -2-(x-17)=34-(-x+25)
-2-x+17=34+x+25
-2 +(-x)+17+(-x)=34+25
[(-x)+(-x)]+[(-2)+17]=29
(-x).2+15=29
(-x).2=29-15
(-x).2=14
(-x)=14:2
(-x)=7
x=-7
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) * ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34-0\\2x=0+6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
\(\left(2019-x\right)\left(3x-12\right)=0\)
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=0+12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy,` x \in {2019; 4}`
p/s: Bài này hnhu mk làm r mà ạ?