`22(x+32)-5=55. 179^0`

`=>22(x+32)-5=55.1`

`=>22(x+32)-5=55`

`=> 22(x+32)=55+5`

`=>22(x+32)=60`

`=>x+32=60:22`

`=>x+32=30/11`

`=>x=30/11-32`

`=>x=-322/11`

`22(x+32)-5=55.179^0`

`=>22(x+32)=55.1+5`

`=>22(x+32)=60`

`=>x+32=60:22`

`=>x+32=30/11`

`=>x=30/11-32`

`=>x=30/11-352/11`

`=>x=-322/11`

Vậy `x=-322/11`

`Yiuby`

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

2^2.(x+3^2)-5=55
           4.(x+9)=55+5=60
     (x+9)=60:4=15
           x=15-9=6
             vậy x=6

\(3^2.5-2^2.7+83\)

\(=9.5-4.7+83\)

= \(45-28+83\)

= \(100\)

a) x+32=5

x=5-32

x=-27

b)x-4=-7

x=-7+4

x=-3

c) 10-(x+12)=5-(22-30)

10-x-12=5-22+30

-2-x=13

x=-2-13

=-15

#H

a,  2 3 x + 5 2 x = 2 5 2 + 2 3 - 33

8x+25x = 33

33x = 33

x = 1

b,  260 : x + 4 = 5 2 3 + 5 - 3 3 2 + 2 2

260:(x+4) = 5.13–3.13

x+4 = 260:26

x+4 = 10

x = 6

c,  720 : [ 41 - 2 x - 5 ] = 2 3 . 5

41–(2x–5) = 720:40

2x–5 = 41–18

2x = 28

x = 14

d,  3 2 - 2 x - 12 + 35 = 5 2 + 279 : 3 2

7(x–12)+35 = 56

7(x–12) = 21

x–12 = 3

x = 15

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