34 - 2x. ( 2x - 6 ) = 0
c) ( 34 - 2x ) . ( 2x - 6 ) = 0
d) ( 2019 - x ) . ( 3x - 12 ) 0
nhanh nha, mik tick cho
`@` `\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}`
Hãy giải các phương trình sau đây :
1, x2 - 4x + 4 = 0
2, 2x - y = 5
3, x + 5y = - 3
4, x2 - 2x - 8 = 0
5, 6x2 - 5x - 6 = 0
6,( x2 - 2x )2 - 6 (x2 - 2x ) + 5 = 0
7, x2 - 20x + 96 = 0
8, 2x - y = 3
9, 3x + 2y = 8
10, 2x2 + 5x - 3 = 0
11, 3x - 6 = 0
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
c) ( 34 - 2x ) . ( 2x - 6 ) = 0
d) ( 2019 - x ) . ( 3x - 12 ) 0
nhanh nha, mik tick cho, giải thik rõ nhé
`@` `\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?
5 (x - 7) = 0
34 (2x - 6) = 0
a) 5.(x-7)=0
=> x-7 = 0
x = 7
b) 34.(2x-6)=0
=> 2x -6 = 0
2x = 6
x = 3
5 (x - 7) = 0
=> x - 7 = 0
=> x = 7
34 (2x - 6) = 0
=> 2x - 6 = 0
=> 2x = 6
=> x = 3
5 (x - 7) = 0
<=> x-7=0
<=> x=7
34 (2x - 6) = 0
<=> 2x-6=0
<=> 2x=6
<=> x=3
Bài 1:Tìm x a,|17-x|-78=99-176-(34-51). b,6+|6-2x|-2x=0
bấm máy tính được luôn đó bạn
nói chứ tính giá trị của trị tuyệt đối rồi đặt 2 trường hợp là x nhỏ hơn 17 và x lớn hơn 17. giải ra nghiệm
ahihi
tìm x:
34(2x - 6 ) = 0
34(2x-6)=0
2x-6=0
2x=6
x=3
=.= hok tốt!!
34(2x - 6) = 0
2x - 6 = 0 : 34
2x - 6 = 0
2x = 0 + 6
2x = 6
x = 6: 2
x = 3
Chúc bạn học tốt!
Thân!
34(2x-6)=0
(2x-6)=0:34
(2x-6) =0
2x =0+6
2x =6
x =6:2
x =3
b) 17x – ( -16x – 37) = 2x + 43
c) -2x –3. (x – 17) = 34 – 2(-x + 25)
d) (-x - 6) (x + 9) = 0
mình gửi lộn
b) 17x – ( -16x – 37) = 2x + 43
-> 33x +37 =2x +43
->31x=6
-> x= 6/31
c) -2x –3. (x – 17) = 34 – 2(-x + 25)
->-5x + 51= 2x-16
->7x= 67
x= 67/7
d) (-x - 6) (x + 9) = 0
Ta có (-x-6)= 0 hoặc (x+9)=0
-> x=-6 hoặc x=-9
b) 17x – ( -16x – 37) = 2x + 43
\(\Leftrightarrow\)17x+16x+37=2x+43
\(\Leftrightarrow\)33x+37-2x-43=0
\(\Leftrightarrow\)31x=6
\(\Leftrightarrow\)x=31/6
c) -2x –3. (x – 17) = 34 – 2(-x + 25)
\(\Leftrightarrow\)-2x-3x-51=34+2x-50
\(\Leftrightarrow\)-5x-51=-16+2x
\(\Leftrightarrow\)-5x-51+16-2x=0
\(\Leftrightarrow\)-7x=35
\(\Leftrightarrow\)x=-5
d) (-x - 6) (x + 9) = 0
\(\hept{\frac{-x-6=0}{x+9=0}}\Leftrightarrow\)\(\hept{\begin{cases}x=-6\\x=-9\end{cases}}\)
Bài 1:
a,-[x+84]+213=-16
b,11-[-53+x]=97
c,/x+10/=105-[-100]
d,/17-x/-78=99-176-[34-51]
e,/x-4/-x+4=0
g,6+/6-2x/-2x=0
h,/8x-9/=/6-7x/
i,/3x+15/-/5-2x/=0
bài 19: tìm x
a) 5 . ( x - 7 ) = 0
b) 25 ( x - 4 ) = 0
c) ( 34 - 2x ) . ( 2x - 6 ) = 0
d) ( 2019 - x ) . ( 3x - 12 ) 0
e) 57 . ( 9x - 27 ) = 0
f) 25 + ( 15 - x ) = 30
g) 43 - ( 24 - x ) = 20
h) 2 . ( x - 5 ) - 17 = 25
i) 3 . ( x + 7 ) - 15 = 27
j) 15 + 4 . ( x - 2 ) = 95
k) 20 - ( x + 14 ) = 5
l) 14 + 3 . ( 5 - x ) = 27
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