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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}`
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: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
a. 5x(x-100)=0
=> TH1 : 5x=0 => x= 0
TH2 : x-100 = 0 => x= 100
Vậy .......
b. 34x (2x-6) =0
TH1: 34x=0 => x= 0
TH2; 2x-6=0 => 2x= 6 => x = 3
Vậy ............
a,5x(x-100)=0
=> 5x=0 hoac x-100=0
=> x=0 hoac x=100
b,34x(2x-6)=0
=> 34x=0 hoac 2x-6=0
=>x=0 hoac x=3
c,3x(x+7)-15=27
phan nay nhieu truong hop lam ban dong nao nha goi y
3x=6hoac21
x+7=7hoac 2
va co 4truong hop
phan con lai ko ro ban viet lai di
\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!
20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
\(\left(2x-6\right)^5+27\left(2x-6\right)^2=0\)
\(\Leftrightarrow\left(2x-6\right)^2\left[\left(2x-6\right)^3+27\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3+27=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=-27\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\2x-6=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{3;\frac{3}{2}\right\}\)