a: \(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left[30-4^2\right]\)

\(=20-14=6\)

b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)

\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)

\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)

c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)

\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)

\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)

d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)

\(=411-\left[\dfrac{110}{5}-4\right]\)

=410-22+4

=410-18

=392

e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)

\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)

\(=450-5\cdot\left[9\cdot8-12\right]+18\)

=468-5*60

=468-300

=168

f:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

290-10.(2018^0+3^5:3^2)

=290-10.(1+243:9)

=290-10.(1+27)

=290-10.28

=290-280

=10

71-50:[5+3.(57-6.7)]

e)x-43=(57-x)-50. 

Suy ra x-43=57-x-50=57-50-x=7-x

Suy ra x-43+x=7. suy ra2x=7+43=50

Nên x=50:2=25

f)-(x-3+84)=(x+70-71)-5.

Suy ra -x+3-84=x-1-5=x-6

Suy ra -x-81=x-6. suy ra -x-81+6=x.suy ra-x-75=x

Suy ra -x-x=75 nên -2x=75 suy ra x=-75/2=-37.5

g)2(x-23)-(x-43)=96-115.

Suy ra 2x-46-x+43=-19,

suy ra 2x-x-46+43=-19,

suy ra x-3=-19

Nên x=-19+3=-16

h)2(x-12)+13=x+(-14)=x-14.

Suy ra 2x-24+13=x-14,

suy ra 2x-11=x-14,

suy ra 2x-11+14=x=2x+3,

suy ra x-2x=3 nên -x=3

vậy x=-3

e)x=25

f)x=-37,5

g)x=-16

h)x=-3

\(\Rightarrow A=\frac{600}{12}+\frac{240}{12}+\frac{100}{12}+\frac{60}{12}+\frac{40}{12}.\)

 \(=\frac{2080}{12}=\frac{520}{3}.\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{7}\right)+.....+\frac{1}{2}\left(\frac{50}{89}-\frac{50}{99}\right)+\frac{1}{99}\)

\(=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{7}+\frac{50}{7}-\frac{50}{8}+.................+\frac{50}{98}-\frac{50}{99}\right)+\frac{1}{99}\)

 \(=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)+\frac{1}{99}\)

Từ A và C ta có \(B=\frac{520}{3}+\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)+\frac{1}{99}\)

                             \(=\left(\frac{520}{3}+\frac{1}{99}\right)+\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)\)

                               \(=\frac{17161}{99}+\frac{1}{2}x\frac{775}{99}\)

                                \(=\frac{17161}{198}+\frac{17161}{99}=\frac{17161}{198}+\frac{34322}{198}=\frac{17161}{66}\)

vậy biểu thức\(B=\frac{17161}{66}\)

Ta chia B thành 2 phần là A và C

Ta có :\(A=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}\)

           \(B=\frac{100}{6x7}+..........+\frac{100}{98x99}+\frac{1}{99}\)

TỪ NHA BN MK LÀM ĐÉN ĐÓ TÍ NỮA MK LÀM TIẾP H MK CÓ VIỆC BN ZÙI

Tui còn chả hiểu nổi cái quy luật của nó như thế nào nữa.

quy luật là gì. mik xem mà ngất luôn

\(A=\dfrac{100}{1\cdot2}+\dfrac{100}{2\cdot3}+\dfrac{100}{3\cdot4}+...+\dfrac{100}{99\cdot100}\)

\(A=100\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

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

\(A=100\cdot\left(1-\dfrac{1}{100}\right)\)

\(A=100\cdot\dfrac{99}{100}\)

A=99