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(x+3).(2x-18)=0

=>x+3=0 hoặc 2x-18 =0

+)x+3=0

x=-3

+)2x-18=0

2x =18

x =9

Vậy x \(\in\)\(_{\left\{-3;9\right\}}\)

2x+35=x-27

2x-x =-35-27

x = -62

Bài 1:

a, 2x + 35 = x - 27

    2x - x = -27 - 35

    x = -62

 b, 2x - 41 = 3x + (-23)

     2x - 3x = -23 + 41

     -x = 18

     x = -18

c, 4 . (x - 3) - 3 . (x - 5) = 45 . (-2) - 34

   4x - 12 - 3x + 15 = -90 - 34

   4x - 3x = -90 - 34 + 12 - 15

   x = -127

Bài 2:

- ( -1234 + 345 - 29 ) - ( 1234 + 135 - 59 )

= 1234 - 345 + 29 - 1234 - 135 + 59

= (1234 - 1234) - (345 + 135) + (29 + 59)

= 0 - 480 + 88

= -392

a)2x+35=x-27

2x-x=-27-35

1x=-62

x=-62:1

X=-62

a: \(\frac{23}{45}\cdot\frac{15}{16}\cdot\frac{32}{23}+0,3\cdot x=5\frac35\cdot7\frac{7}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)

=>\(\frac{12}{16}\cdot\frac{15}{45}+0,3x=\frac{28}{5}\cdot\frac{91}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)

=>\(0,3x+\frac{1}{12}=\frac{91}{13}\cdot\frac{2}{12}\cdot\frac{28}{27}=7\cdot\frac16\cdot\frac{28}{27}=\frac{7\cdot14}{3\cdot27}=\frac{98}{81}\)

=>\(0,3x=\frac{98}{81}-\frac{1}{12}=\frac{365}{324}\)

=>\(x=\frac{365}{324}:0,3=\frac{365}{324\cdot0,3}=\frac{365}{97,2}=\frac{1825}{486}\)

b: \(\left\lbrack\left(35\frac57+2\frac34\right)-5\frac57+\frac14\right\rbrack:11+x=3\)

=>\(\left\lbrack35+\frac57+2+\frac34-5-\frac57+\frac14\right\rbrack:11+x=3\)

=>(32+1):11+x=3

=>33/11+x=3

=>x+3=3

=>x=0

\(a.2x+35=x-27\)

\(\Leftrightarrow2x-x=-27-35\)

\(\Rightarrow x=-62\). Vậy \(x=-62\)

\(b.2x-41=3x+\left(-23\right)\)

\(\Leftrightarrow2x-3x=-23+41\)

\(\Leftrightarrow-x=18\)

\(\Rightarrow x=-18\). Vậy \(x=-18\)

\(c.4\left(x-3\right)-3\left(x+5\right)=45.\left(-2\right)-34\)

\(\Leftrightarrow4x-12-3x-15=-124\)

\(\Leftrightarrow4x-3x=-124+12+15\)

\(\Rightarrow x=-97\). Vậy \(x=-97\)

\(d.\left(x+3\right)\left(2x-18\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x-18=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\x=9\end{cases}}}\). Vậy \(x=-3;9\)

a,x.3\(\dfrac{1}{4}\)+(\(\dfrac{-7}{6}\)).x-1\(\dfrac{2}{3}\)=\(\dfrac{5}{12}\)

\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x-\(\dfrac{5}{3}\)=\(\dfrac{5}{12}\)

\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x=\(\dfrac{25}{12}\)

\(\Leftrightarrow\)\(\dfrac{25}{12}\)x=\(\dfrac{25}{12}\)

\(\Leftrightarrow\)x=1

Vậy x=1

b,\(\dfrac{2}{5}\)+\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)

\(\Leftrightarrow\)\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)-\(\dfrac{2}{5}\)

\(\Leftrightarrow\)\(\dfrac{9}{5}\)x-\(\dfrac{111}{50}\)=\(\dfrac{-57}{10}\)

\(\Leftrightarrow\)\(\dfrac{9}{5}\)x=\(\dfrac{-87}{25}\)

\(\Leftrightarrow\)x=\(\dfrac{-29}{15}\)

Vậy x=\(\dfrac{-29}{15}\)

c,\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)+\(\dfrac{5}{9}\)=\(\dfrac{23}{27}\)

\(\Leftrightarrow\)\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)=\(\dfrac{8}{27}\)

\(\Leftrightarrow\)\(\dfrac{7}{18}\)+\(\dfrac{28}{27}\)x=\(\dfrac{8}{27}\)

\(\Leftrightarrow\)\(\dfrac{28}{27}\)x=\(\dfrac{-5}{54}\)

\(\Leftrightarrow\)x=\(\dfrac{-5}{56}\)

Vậy x=\(\dfrac{-5}{56}\)

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