giúp mik nhé

đi mà giúp đi

e)175:x=345:43

3,4:x=6,8:43

3,4:x=5,1

x=3,4:5,1

x=23

f)x:45 =258:54

x:45=52

x=52×45

x=2

g)(x×0,25+2012)×2013=(50+2012)×2013g)

(x×0,25+2012)×2013=2062×2013

x×0,25+2012=(2062×2013):2013

x×0,25+2012=2062

x×0,25=2062−2012

x×0,25=50

x=50:0,25

x=200

h)(x−12)×53=74−12h)

(x−12)×53=54

x−12=54:53

x−12=34

x=34+12

x=54

e: \(\dfrac{17}{5}:x=\dfrac{34}{5}:\dfrac{4}{3}=\dfrac{34}{5}\cdot\dfrac{3}{4}=\dfrac{102}{20}=\dfrac{51}{10}\)

\(\Leftrightarrow x=\dfrac{17}{5}:\dfrac{51}{10}=\dfrac{17}{5}\cdot\dfrac{10}{51}=\dfrac{10}{5}\cdot\dfrac{17}{51}=2\cdot\dfrac{1}{3}=\dfrac{2}{3}\)

f: \(x:\dfrac{4}{5}=\dfrac{25}{8}:\dfrac{5}{4}=\dfrac{25}{8}\cdot\dfrac{4}{5}=\dfrac{100}{40}=\dfrac{5}{2}\)

\(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{4}{5}=\dfrac{4}{2}=2\)

g: \(\left(0.25x+2012\right)\cdot2013=\left(50+2012\right)\cdot2013\)

\(\Leftrightarrow0.25x+2012=50+2012\)

\(\Leftrightarrow0.25x=50\)

hay x=200

h: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{3}=\dfrac{7}{4}-1=\dfrac{3}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{3}{4}:\dfrac{5}{3}=\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{9}{20}+\dfrac{1}{2}=\dfrac{9}{20}+\dfrac{10}{20}=\dfrac{19}{20}\)

\(4\left(x-1\right)-3\left(x-2\right)=-5\)

\(\Leftrightarrow4x-4-3x+6=-5\)

\(\Leftrightarrow x=-5+4-6\)

\(\Leftrightarrow x=-7\)

Vậy x=-7

Ta có: 4(x-1) - 3(x-2) = -5

          (4x-4) - (3x-6) = -5

          4x - 4 - 3x + 6 = -5 

     (4x - 3x) + (-4+6) = -5

                       x + 2  = -5

                             x  = -5 - 2

                              x = -7

                         Vậy x = -7

giúp mik nhanh với ah mik cần gấp

0,25 x 3 + 1 : 4 x 7

= 0,25 (3+7)

= 0,25 x 10

= 2,5

X x 1.2 + X x 1.8 = 45

X [1(2+8)] = 45

X [1 x 10] = 45

X x 10 = 45

        X = 45 : 10

        X = 4,5

A.\(\left(\frac{4}{7}.x-1\right).3=\frac{1}{21}\)

\(\frac{4}{7}.x-1=\frac{1}{21}:3\)

\(\frac{4}{7}.x-1=\frac{1}{21}.3\)

\(\frac{4}{7}.x-1=\frac{1}{7}\)

\(\frac{4}{7}.x=1+\frac{1}{7}\)

\(\frac{4}{7}.x=\frac{8}{7}\)

\(x=\frac{8}{7}:\frac{4}{7}\)

\(x=\frac{8}{7}.\frac{7}{4}\)

\(x=2\)

Vậy x=2

C.\(\frac{2}{3}.x-\frac{1}{3}.x=\frac{5}{9}\)

\(\left(\frac{2}{3}-\frac{1}{3}\right).x=\frac{5}{9}\)

\(\frac{1}{3}.x=\frac{5}{9}\)

\(x=\frac{5}{9}:\frac{1}{3}\)

\(x=\frac{5}{9}.3\)

\(x=\frac{5}{3}\)

Vậy \(x=\frac{5}{3}\)

D. \(\left(x-2\frac{1}{4}\right).\left(\frac{-2}{3}\right).50\%=2\frac{5}{6}\)

\(x.\frac{-2}{3}-\frac{9}{4}.\frac{-2}{3}=\frac{17}{6}:\frac{1}{2}\)

\(x.\frac{-2}{3}-\frac{-3}{2}=\frac{17}{6}.2\)

\(x.\frac{-2}{3}+\frac{3}{2}=\frac{17}{3}\)

\(x.\frac{-2}{3}=\frac{17}{3}-\frac{3}{2}\)

\(x.\frac{-2}{3}=\frac{34}{6}-\frac{9}{6}\)

\(x.\frac{-2}{3}=\frac{25}{6}\)

\(x=\frac{25}{6}:\frac{-2}{3}\)

\(x=\frac{25}{6}.\frac{3}{-2}\)

\(x=\frac{25}{-4}\)

Vậy \(x=\frac{25}{-4}\)

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)

\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)

a: \(-\dfrac{2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\)

=>\(\dfrac{4}{5}x=\dfrac{3}{5}+\dfrac{2}{5}=1\)

=>\(x=1:\dfrac{4}{5}=\dfrac{5}{4}\)

b; \(-\dfrac{3}{7}-\dfrac{4}{7}:x=-2\)

=>\(\dfrac{4}{7}:x+\dfrac{3}{7}=2\)

=>\(\dfrac{4}{7}:x=2-\dfrac{3}{7}=\dfrac{11}{7}\)

=>\(x=\dfrac{4}{7}:\dfrac{11}{7}=\dfrac{4}{11}\)

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)