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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

tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

k. \(\left(2x+\dfrac{3}{5}\right)\cdot2-\dfrac{9}{25}=0\\ \Leftrightarrow\left(2x+\dfrac{3}{5}\right)\cdot2=\dfrac{9}{25}\\ \Leftrightarrow2x+\dfrac{3}{5}=\dfrac{9}{50}\\ \Leftrightarrow2x=\dfrac{-21}{50}\\ \Leftrightarrow x=\dfrac{-21}{100}\)
l. \(3\left(3x-\dfrac{1}{2}\right)\cdot3+\dfrac{1}{9}=0\\ \Leftrightarrow\left(3x-\dfrac{1}{2}\right)\cdot9=-\dfrac{1}{9}\\ \Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{81}\\ \Leftrightarrow3x=\dfrac{79}{162}\\ \Leftrightarrow x=\dfrac{79}{486}\)

a)\(2\left(x-\frac{1}{2}\right)^3-\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x=1\)
b)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
c)\(\left(2n+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2n+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2n+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2=\left(-\frac{3}{5}\right)^2\)
\(\Rightarrow\hept{\begin{cases}2n+\frac{3}{5}=\frac{3}{5}\\2n+\frac{3}{5}=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}n=0\\n=-\frac{3}{5}\end{cases}}\)
Vậy n=0;-3/5
d)\(3\left(3n-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\left(3n-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\left(3n-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(3n-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow n=\frac{1}{18}\)

(x - 2)(2x - 6) = 0
=> x - 2 = 0 hoac 2x - 6 = 0
=> x = 2 hoac x = 3
vay_
(3x + 9)(1 - 3x) = 0
=> 3x + 9 = 0 hoac 1 - 3x = 0
=> x = -3 hoac x = 1/3
vay_
|2 - x| + 2 = x
=> |2 - x| = x - 2
=> 2 - x = x - 2 hoac 2 - x = 2 - x
=> -2x = -4 hoac x thuoc tap hop rong
=> x = 2
\(a,\left(x-2\right).\left(2x-6\right)=0\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(b,\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow3\left(x+3\right).\left(1-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
\(c,\left(x^2+1\right).\left(81-x^2\right)=0\Leftrightarrow\left(x^2+1\right).\left(9-x\right).\left(9+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=0\\9+x=0\end{cases}}\) ( vì \(x^2+1\ne0\forall x\)) \(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)

a)4x+4-3x+1=14
x+5=14
x=11
b)trường hợp 1 x2-9=0
x2=9
->x=3;-3
-trường hợp 2: x+2=0
x=-2
c)-th1:x2+9=0
x2=-9
->x rỗng
d)xy+2x-y-2=0
(xy-y)+(2x-2)=0
y(x-1)+2(x-1)=0
(y+2)(x-1)=0
th1: y+2=0
y=-2
th2:x-1=0
x=1
(th1: trường hợp 1)

Ta có : 7(x - 1) + 2x(x - 1) = 0
<=> (2x + 7)(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-7\\x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=1\end{cases}}\)

\(\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{3}\\2x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}\\2x=-\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}:2=-\dfrac{1}{12}\\x=-\dfrac{5}{6}:2=-\dfrac{5}{12}\end{matrix}\right.\)
\(\dfrac{1}{9}-\left(2x+\dfrac{1}{2}\right)^2=0\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)=\dfrac{1}{9}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=\dfrac{1}{3}\\2x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{6}\\2x=\dfrac{1}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)