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( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 40
x + ( 1 + 2 + 3 + 4 + 5 ) = 40
x + 15 = 40
x = 25
( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 40
x + ( 1 + 2 + 3 + 4 + 5 ) = 40
x + 15 = 40
x = 25
a: =>y+1,2=(5,34-10,34):5=-1
=>y=-2,2
b: =>x(x+1)/2=111a
=>x(x+1)=222a
=>\(x\in\varnothing\)
A) \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)
\(x=\dfrac{22}{15}\)
b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)
C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)
a, \(x:3\dfrac{5}{6}=2\dfrac{8}{15}\)
\(x:\dfrac{23}{6}\) = \(\dfrac{38}{15}\)
\(x=\dfrac{38}{15}\times\dfrac{23}{6}\)
\(x=\dfrac{437}{45}\)
b, \(\dfrac{12}{15}:x+\dfrac{4}{3}=3\dfrac{2}{3}\)
\(\dfrac{12}{15}:x+\dfrac{4}{3}=\dfrac{11}{3}\)
\(\dfrac{12}{15}:x=\dfrac{11}{3}-\dfrac{4}{3}\)
\(\dfrac{12}{15}:x=\dfrac{7}{3}\)
x= \(\dfrac{12}{15}:\dfrac{7}{3}=\dfrac{12}{35}\)
\(c)x:\dfrac{5}{4}=\dfrac{9}{5}+\dfrac{1}{2}\)
\(x:\dfrac{5}{4}=\dfrac{23}{10}\)
\(x=\dfrac{23}{10}\times\dfrac{5}{4}\)
\(x=\dfrac{23}{8}\)
a, 2x3x4x8x25x125= (2x3)x(4x25)x(8x125)= 6x100x1000=600000
b,42/5 +56/9+23/4+3/5+1/3+1/4= (42/5+3/5)+(23/4+1/4)+(56/9+1/3)=9+6+59/9=
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
1. a
\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)
\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)
\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)
1.b
\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)
\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)
2.a
\(\dfrac{4}{5}+x=\dfrac{11}{10}\)
\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)
2.b
\(x-\dfrac{3}{4}=\dfrac{5}{7}\)
\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)
\(x\times2\dfrac{3}{4}=\dfrac{5}{2}:\dfrac{2}{3}\\ =>x\times\dfrac{11}{4}=\dfrac{15}{4}\\ =>x=\dfrac{15}{4}:\dfrac{11}{4}\\ =>x=\dfrac{15}{11}\)