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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 !
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
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)\(3\left(3n-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\left(3n-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\left(3n-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(3n-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow n=\frac{1}{18}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
![](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) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
Đổi: 675km = 67 500 000cm
Trên bản đồ tỉ lệ 1:2 500 000 quãng đường dài là:
67 500 000 : 2 500 000 = 27 (cm)
Đáp số: 27 cm
Xin lỗi nha