Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

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 \(x = \frac{{ - 88}}{{15}}\).

a/  => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)

=> \(\dfrac{1}{x}=\dfrac{2}{5}\)

=> x = 5/2

b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)

=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)

=> \(x=\dfrac{2}{5}\)

c/ => | x + 1| = 10/21

=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)

d/ => \(5x+5=6x-3\)

=> x = 8

mọi người ơi giúp mk đi mk dốt toán lém

Bài 1: 

a: Số phần tử của tập hợp A là:

50-11+1=40

b: Số phần tử của tập hợp B là:

\(\left(100-0\right):10+1=11\)

c: Tập hợp C có 1 phần tử

d: Tập hợp D có : \(\left(31-5\right):2+1=14\)

e: Tập hợp E có 5 phần tử

f: Tập hợp F có vô số phần tử

a)

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)

\(x+x+2+x+4+...+x+98=0\)

\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)

\(50x+100.49:2=0\)

\(50x+49.50=0\)

\(50x=0-49.50\)

\(50x=-2450\)

\(x=-2450:50\)

\(x=-49\)

b)

\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)

\(x+x+x+...+x-5-4-3-...+11+12=99\)

\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)

\(18x+63=99\)

\(18x=99-63\)

\(18x=36\)

\(x=36:18\)

\(x=2\)

giúp mình với, mình đang vội!

a) x + (x + 2) + (x + 4) + ... + (x + 98) = 0

x + x + 2 + x + 4 + ... + x + 98 = 0

50x + (98 + 2).[(98 - 2) : 2 + 1]:2 = 0

50x + 100 .49 : 2 = 0

50x + 49.50 = 0

50x = 0 - 49.50

50x = -2450

x = -2450 : 50

x = -49

b) (x - 5) + (x - 4) + (x - 3) + ... + (x + 11) + (x + 12) = 99

x + x + x + ... + x - 5 - 4 - 3 - ... + 11 + 12 = 99

18x + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 99

18x + 63 = 99

18x = 99 - 63

18x = 36

x = 36 : 18

x = 2

c: Ta có: 11+x:5=13

\(\Leftrightarrow x:5=2\)

hay x=10

d: Ta có: \(13+2\left(x+1\right)=15\)

\(\Leftrightarrow2x+2=2\)

\(\Leftrightarrow2x=0\)

hay x=0

e: Ta có: 2x+21=41

\(\Leftrightarrow2x=20\)

hay x=10

f: Ta có: \(12+3\left(x-2\right)=60\)

\(\Leftrightarrow3\left(x-2\right)=48\)

\(\Leftrightarrow x-2=16\)

hay x=18

g: Ta có: \(24x-11\cdot13=11\cdot11\)

\(\Leftrightarrow24x=11\cdot24\)

hay x=11

h: Ta có: \(17-\left(x-4\right):2=3\)

\(\Leftrightarrow\left(x-4\right):2=14\)

\(\Leftrightarrow x-4=28\)

hay x=32

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

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

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a: =>x+16=-17

hay x=-33