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

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3(3x-1/2)^3+1/9=0

`Answer:`

a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)

b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{9}:3\)

\(\Leftrightarrow x=\frac{1}{27}\)

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

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!!

g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)

\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)

c. \(3x+2x=\frac{5}{8}\times\frac{4}{3}\)

\(\Leftrightarrow x\left(3+2\right)=\frac{5.4}{8.3}\)

\(\Leftrightarrow5x=\frac{5.1}{2.3}\)

\(\Leftrightarrow5x=\frac{5}{6}\)

\(\Leftrightarrow x=\frac{5}{6}\div5\)

\(\Leftrightarrow x=\frac{1}{6}\)

ủng hộ mk nha bn

bạn trả lời giúp

tk cho mk nhé mình đang bị âm nek