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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 !
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
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
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
a: \(\Leftrightarrow x\cdot\dfrac{1}{2}=\dfrac{1}{4}\cdot\dfrac{-7}{2}+\dfrac{5}{3}=\dfrac{19}{24}\)
hay x=19/12
b: \(\Leftrightarrow\left(\dfrac{45}{11}-3x\right)\cdot\dfrac{11}{5}=\dfrac{21}{5}\)
\(\Leftrightarrow-3x+\dfrac{45}{11}=\dfrac{21}{11}\)
=>-3x=-24/11
=>x=8/11
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
d: \(\Leftrightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{4}{5}\\2x+\dfrac{3}{5}=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)