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Bài 4:

a) \(\dfrac{x}{2}=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Rightarrow x=2\)

Vậy: \(x=2\)

b) \(-\dfrac{1}{5}=\dfrac{2}{x}\)

\(\Rightarrow x=\dfrac{-5.2}{1}=-10\)

Vậy: \(x=-10\)

c) \(\dfrac{x}{5}=\dfrac{5}{x}\)

\(\Rightarrow x^2=25\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-5\right\}\)

\(\dfrac{x}{2}=\dfrac{2}{4}\\ =>x=\dfrac{2.2}{4}=1\)

\(\dfrac{-1}{5}=\dfrac{2}{x}\\ =>x=\dfrac{2.5}{-1}=-10\)

\(\dfrac{x}{5}=\dfrac{5}{x}\\ =>x^2=25\\ x=5;x=-5\)

a) 2

b) 5,-5

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

d) 

+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\)  (Nhận)

+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)

+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)

+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)

  Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\) 

a, \(\left(\dfrac{1}{2}+\dfrac{4}{7}\right):x=\dfrac{-3}{4}\)

\(\dfrac{15}{14}:x=\dfrac{-3}{4}\)

=> x= \(\dfrac{-7}{10}\)

b, 0,5:x-\(1\dfrac{3}{4}\)= 25%

0,5:x-\(\dfrac{7}{4}=\dfrac{1}{4}\)

0,5:x = 2

=> x = \(\dfrac{1}{4}\)

a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)

=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)

=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)

=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)

b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)

=>\(x\cdot\dfrac{-1}{6}=0,75\)

=>\(x=-0,75\cdot6=-4,5\)

`@` ` \text {Ans}`

`\downarrow`

`a,`

`1/4+3/4*x=3/2-x`

`=> 1/4 + 3/4x - 3/2 + x = 0`

`=> (1/4 - 3/2) + (3/4x + x) = 0`

`=> -5/4 + 7/4x = 0`

`=> 7/4x = 5/4`

`=> x = 5/4 \div 7/4`

`=> x = 5/7`

Vậy, `x=5/7`

`b,`

`3/5*x-1/4=1/10*x-1/2`

`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`

`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`

`=> 1/2x + 1/4 = 0`

`=> 1/2x = -1/4`

`=> x = -1/4 \div 1/2`

`=> x = -1/2`

Vậy, `x=-1/2`

`c,`

`3x-3/5=x-1/4`

`=> 3x - 3/5 - x + 1/4 = 0`

`=> (3x - x) - (3/5 - 1/4) = 0`

`=> 2x - 7/20 = 0`

`=> 2x = 0,35`

`=> x = 0,35 \div 2`

`=> x = 7/40`

Vậy, `x=7/40`

`d,`

`3/2*x-2/5=1/3*x-1/4`

`=>  3/2x - 2/5 - 1/3x + 1/4 = 0`

`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`

`=> 7/6x - 3/20 = 0`

`=> 7/6x = 3/20`

`=> x = 3/20 \div 7/6`

`=> x = 9/70`

Vậy, `x=9/70`

`@` `\text {Kaizuu lv uuu}`