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a: =>|x+3/4|=2+1/5=11/5
=>x+3/4=11/5 hoặc x+3/4=-11/5
=>x=29/20 hoặc x=-59/20
b: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
c: =>|2x-1/3|=1/6
=>2x-1/3=1/6 hoặc 2x-1/3=-1/6
=>2x=1/2 hoặc 2x=1/6
=>x=1/4 hoặc x=1/12
e: =>x+2/3=0 hoặc -2x-3/5=0
=>x=-2/3 hoặc x=-3/10
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) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
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!!
Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
a) \(\left\{\left(\dfrac{3}{5}\right)^2-\left(\dfrac{2}{5}\right)^2\right\}x=\left(\dfrac{1}{5}\right)^3\)
\(\left(\dfrac{9}{25}-\dfrac{4}{25}\right)x=\dfrac{1}{125}\)
\(\dfrac{1}{5}x=\dfrac{1}{125}\)
\(x=\dfrac{1}{25}\)
b) Ý này đề bài rõ hơn đi ạ
c)\(\left(3x-2^2\right)=9\)
\(3x-4=9\)
\(3x=13\)
\(x=\dfrac{13}{3}\)
d)\(\left(x+\dfrac{1}{3}\right)-4=-1\)
\(x+\dfrac{1}{3}=3\)
\(x=\dfrac{8}{3}\)
a, \(\left[\left(\dfrac{3}{2}\right)^2-\left(\dfrac{2}{5}\right)^2\right]x=\left(\dfrac{1}{5}\right)^3\)
\(\Leftrightarrow\left(\dfrac{9}{4}-\dfrac{4}{10}\right)x=\dfrac{1}{125}\)
\(\Leftrightarrow\dfrac{1}{5}x=\dfrac{1}{125}\)
\(\Leftrightarrow x=\dfrac{1}{125}.5=\dfrac{1}{25}\)
b, Ghi lại đề
c, \(\left(3x-2^2\right)=9\)
\(\Leftrightarrow3x-4=9\)
\(\Leftrightarrow x=\dfrac{9+4}{3}=\dfrac{13}{3}\)
d, \(\left(x+\dfrac{1}{3}\right)-4=-1\)
\(\Leftrightarrow x=-1+4-\dfrac{1}{3}=3-\dfrac{1}{3}=\dfrac{8}{3}\)