a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0

a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran

c, Do (x - \(\dfrac{1}{2}\))(x + \(\dfrac{3}{4}\)) < 0
=> x - \(\dfrac{1}{2};x+\dfrac{3}{4}\) khác dấu
Có x - \(\dfrac{1}{2}< x+\dfrac{3}{4}\Rightarrow x-\dfrac{1}{2}< 0;x+\dfrac{3}{4}>0\)
=> \(\left[{}\begin{matrix}x-\dfrac{1}{2}< 0\\x+\dfrac{3}{4}>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>-\dfrac{3}{4}\end{matrix}\right.\)<=> -\(\dfrac{3}{4}< x< \dfrac{1}{2}\)
Do x \(\in Z\Rightarrow x=0\)
@Mai Tran

1: =>x=8-35=-27

2: =>15-4+x=6

=>x+11=6

hay x=-5

3: =>-30+25-x=-1

=>x+5=1

hay x=-4

4: =>x-(-13)=-8

=>x+13=-8

hay x=-21

5: =>x-29-17+38=-9

=>x-8=-9

hay x=-1

(12000+X):376=X

12000+X=376 x X

12000=376 x X - X

12000=375 x X

X=12000:375= 32

1. a) 17.85 + 15.17 - 120

= 17.( 85 + 15) - 120

= 17. 100 -120

= 1700 - 120

= 1580

b) 20 - [ 30 - (5 - 1)² ]

= 20 - [ 30 - 4² ]

= 20 - [ 30 - 16 ]

= 20 - 14

= 6

2. a) 70 - 5.(x - 3) = 45

              5.(x - 3) = 70 - 45

              5.(x -3) = 25

                 x - 3 = 25 : 5

                 x - 3 = 5

                 x = 5 + 3

                 x = 8 

b) 10 + 2.x = 4⁵ : 4³

    10 + 2.x = 4²

    10 + 2.x = 16

           2.x = 16 - 10

          2.x = 6

             x = 6 : 2

            x =3

1)a:17.85+15.17-120=17.(85+15)-120=17.100-120=1700-120=1580

b: 20-[30-(5-1)^2 = 20-[30-4^2]=20-(30-16)=20-15=5

2)

a:  70-5(x-3)=45

   5(x-3) = 70-45=25

   x-3 = 25:5=5

   x= 5+3=8

b:  10+2x = 4^5:4^3

   10+2x = 4^2

10+2x = 16

2x = 16-10=6

x = 6:2=3

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

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

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

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

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...