![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu1
(3x-1).(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
=>x=1/3 =>x=0
câu2
1/4+1/3 :(2x-1)=5
=> 1/3:(2x-1)=19/4
=>2x-1 =57/4
=>2x=61/4
=>x=61/8
còn hai câu sau bn ghi đề mik ko hỉu
1.
a)(3x-1)(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
3x=0+1 x=0:1/2:5
x=1/3 x=0
Vậy x=1/3 hoặc x=0
b)1/4+1/3:(2x-1)=5
1/3:(2x-1)=5-1/4=20/4-1/4=19/4
2x-1=1/3:19/4=1/3*4/19=4/57
2x=4/57+1=4/57+57/57=61/57
x=61/57:2=61/57*1/2=61/114
Vậy x=61/114
c)(2x+2/5)2-9/25=0=02-9/25
=>2x+2/5=0
2x=0-2/5
x=-2/5:2=-2/5*1/2
x=-1/5
Vậy x=-1/5
d)(3x-1/2)3+1/9=0=03+1/9
=>3x-1/2=0
3x=0+1/2
x=1/2:3=1/2*1/3
x=1/6
Vậy x=1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x-3/5)^2-9/25=0
(2x-3/5)^2=9/25
2x-3,5=3/5
2x=3/5+3/5
x=6/5:2=3/5
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Để \(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)=> 3x-1=0 hoặc \(-\frac{1}{2}x+5=0\)
=> x= \(\frac{1}{3}\) hoăc \(x=10\)
b)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=5\) => \(\frac{1}{3}:\left(2x-1\right)=5-\frac{1}{4}=\frac{19}{4}=>2x-1=\frac{1}{3}:\frac{19}{4}=\frac{4}{57}=>x=\frac{61}{114}\)
c) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0=>\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)\(=>2x+\frac{3}{5}\in\left\{\pm\frac{3}{5}\right\}=>2x\in\left\{0;\frac{-6}{5}\right\}=>x\in\left\{0;\frac{-3}{5}\right\}\)
d) Xem lại đề
a) để (3x-1).(\(-\dfrac{1}{2}x+5\))=0
=> 3x-1 hoặc \(-\dfrac{1}{2}x+5\) =0
TH1 : 3x-1=0
3x = 0+1=1
x = 1:3 = \(\dfrac{1}{3}\)
TH2 : \(-\dfrac{1}{2}x+5\)= 0
\(-\dfrac{1}{2}x\)= 0 -5 = -5
x= -5 : \(-\dfrac{1}{2}\)
x= 10
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ (3x - 1).(1/2.5) = 0 => 3x - 1 = 0 => 3x = 1 => x = 1/3
b/ 1/4 + 1/3 : (2x - 1) = 5 => 1/3 : (2x - 1) = 19/4 => 2x - 1 = 4/57 => 2x = 61/57 => x = 61/114
c/ (2x + 2/5)2 - 9/25 = 0 => (2x + 2/5)2 = 9/25 => 2x + 2/5 = 3/5 => 2x = 1/5 => x = 1/10
hoặc 2x + 2/5 = -3/5 => 2x = -1 => x = -1/2
Vậy x = {1/10 ; -1/2}
d/ (3x - 1/2)3 + 1/9 = 0 => (3x - 1/2)3 = -1/9 => 3x - 1/2 = -1/3 => 3x = 1/6 => x = 1/18
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(\(\dfrac{2x}{9}\)+1):(-5)=\(\dfrac{-1}{45}\)
<=>\(\dfrac{2x}{9}\)+1=(\(\dfrac{-1}{45}\))*(-5)
<=>\(\dfrac{2x}{9}\)+1=\(\dfrac{1}{9}\)
<=>\(\dfrac{2x}{9}\)=\(\dfrac{1}{9}\)-1
<=>\(\dfrac{2x}{9}\)=\(\dfrac{1}{9}\)-\(\dfrac{9}{9}\)
<=>\(\dfrac{2x}{9}\)=\(\dfrac{-8}{9}\)
<=>2x=\(\dfrac{-8}{9}\)*9
<=>2x=-8
<=>x=-8:2
<=>x=-4
b)vì x^2 >=0
Mà (x^2-2).(x^2-25)<0
suy ra x^2 -2 là số âm và x^2-25 là số dương
hoặc x^2-25 là số âm và x^2-2 là số dương
mà x^2-2>x^2-25 (2<25) nên x^2-25 là số âm và x^2-2 là số dương
suy ra x thuộc vào tập hợp 2;3;4
vậy x=2 hoặc x=3 hoặc x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)^3=27\)
\(\left(x+1\right)^3=3^3\)
\(\Rightarrow x+1=3\)
\(x=2\)
\(\left(x+1\right)^3=27\)
\(< =>\left(x+1\right)^3=3.3.3=3^3\)
\(< =>x+1=3< =>x=3-1=2\)
\(\left(2x+3\right)^3=9.81\)
\(< =>\left(2x+3\right)^3=9.9.9\)
\(< =>\left(2x+3\right)^3=9^3\)
\(< =>2x+3=9< =>2x=6\)
\(< =>x=\frac{6}{2}=3\)
( 2x + 3/5 )2 - 9/25 = 0
( 2x + 3/5 )2 = 0 + 9/25
( 2x + 3/5 )2 = 9/25
( 2x + 3/5 )2 = ( 3/5 )2
=> 2x + 3/5 = 3/5
=> 2x = 3/5 - 3/5
=> 2x = 0
=> x = 0 : 2
=> x = 0
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\pm\frac{3}{5}\right)^2\)
\(2x+\frac{3}{5}=\pm\frac{3}{5}\)
TH1: TH2:
\(2x+\frac{3}{5}=\frac{3}{5}\) \(2x+\frac{3}{5}=-\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\) \(2x=-\frac{3}{5}-\frac{3}{5}\)
\(2x=0\) \(2x=-\frac{6}{5}\)
\(x=0:2\) \(x=-\frac{6}{5}:2\)
\(x=2\) \(x=-\frac{3}{5}\)