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\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x+81=5\)
\(-19x=5-81\)
\(-19x=-76\)
\(x=-76:\left(-19\right)\)
\(x=4\)
\(Vậyx=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x-30-24x=100\)
\(30x-6x-24x+60-30=100\)
\(0x+30=100\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn đề bài.
\(c,-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-1-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{11}{2}x-\frac{2}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{2}{3}+\frac{5}{6}=\frac{3}{2}x+\frac{11}{2}x\)
\(-\frac{4}{6}+\frac{5}{6}=\frac{14}{2}x\)
\(\frac{1}{6}=7x\)
\(x=\frac{1}{6}:7\)
\(x=\frac{1}{6}.\frac{1}{7}\)
\(x=\frac{1}{42}\)
\(Vậyx=\frac{1}{42}\)
\(d,-3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(-3x+\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(-3x-5x+\frac{3}{2}-3=-x+\frac{1}{5}\)
\(-8x+\frac{3}{2}-\frac{6}{2}=-x+\frac{1}{5}\)
\(-8x-\frac{3}{2}=-x+\frac{1}{5}\)
\(-\frac{3}{2}-\frac{1}{5}=-x+8x\)
\(\frac{15}{10}-\frac{2}{10}=7x\)
\(7x=\frac{13}{10}\)
\(x=\frac{13}{10}:7\)
\(x=\frac{13}{10}.\frac{1}{7}\)
\(x=\frac{13}{70}\)
\(Vậyx=\frac{13}{70}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{-12(x-5)+7(3-x)=5 }\)
\(-12x+60+21-7x=5\)
\(-12x-7x=5-21-60\)
\(-19x=-76\)
\(x=-76:\left(-19\right)\)
\(x=4\)
\(\text{ 30(x+2)-6(x-5)-24x=100}\)
\(30x+60-6x+30-24x=100\)
\(30x-6x-24x=100-30-60\)
\(0=10\)
\(\Rightarrow x\)ko tồn tại
\(\text{(x+1-5)+7(3-x)=5 }\)
\(x+1-5+21-7x=5\)
\(x-7x=5-21+5-1\)
\(-6x=-12\)
\(x=\left(-12\right):\left(-6\right)\)
\(x=2\)
\(\text{(x+1)+(x+3)+(x+5)+........+(x+99)=0}\)
\(\text{(x+1)+(x+3)+...+(x+99)=0}\)
tổng các số hang là\(\frac{\left(99+1\right)}{2}=50\)(số hạng)
=>\(\text{(x+1)+(x+3)+...+(x+99)=0}\)<=> \(\text{50.x+(1+3+5+..+99)=0}\)
<=>\(\text{50.x+(99+1)}\)\(.\frac{50}{2}=0\)<=> \(\text{50.x+2500=0=}\)>\(x=\frac{-2500}{50}=-50\)
chúc bạn học tốt
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a)
x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
b)1 + 2 + 3 +...+ x = 210
(x + 1) . x : 2=210
x . (x + 1) = 420
x . x + x = 420
x . x = 420 - x
Ta thấy:
20 . 20 = 420 - 20
Vậy x = 20
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm ơn giải nha giùm mik, mik đang cần gấp nha! Cảm ơn mọi người .
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a, Ta có :\(-30+\left(25-x\right)=-1\)
\(\Leftrightarrow\left(-30\right)+25-x=-1\)
\(\Leftrightarrow25-x=\left(-1\right)-\left(-30\right)\)
\(\Leftrightarrow25-x=29\\ \Leftrightarrow x=25-29\)
\(\Leftrightarrow x=\left(-4\right)\)
Vậy \(x=-4\)
b,Ta có :\(\left(x+5\right)+\left(x-9\right)=x+2\)
\(\Leftrightarrow x+5+x-9=x+2\)
\(\Leftrightarrow2.x+\left(5-9\right)=x+2\)
\(\Leftrightarrow\) \(2.x+\left(-4\right)=x+2\)
\(\Leftrightarrow2.x-x=4+2\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x+(x+1)+(x+2)+(x+3)+.......+(x+30)=1240
\(\Leftrightarrow\left(x+x+x.+...x\right)+\left(1+2+3...+30\right)=1240\)
\(\Rightarrow30x+465=1240\)
\(\Rightarrow30x=1240-465=775\)
\(\Rightarrow30x=775\)
\(V\text{ậy}x=\frac{155}{6}\)
1+2+3+.....+x=210
\(\left(1+x\right).x=210\)
\(\Rightarrow x=14\)
x+(x+1)+(x+2)+...+(x+30)=1240
=>x+x+1+x+2+...+x+30=1240
=>(x+x+x+...+x)+(1+2+...+30)=1240
=>31x+[(30-1):1+1] . (30+1) :2=1240
=>31x+30.31:2=1240
=>31x+15.31=1240
=>31(x+15)=1240
=>x+15=1240:31=40
=>x=40-15=25
1+2+3+...+x=210
=>[(x-1):1+1]. (x+1) : 2= 210
=>x.(x+1):2=210
=>x(x+1)=210.2=420
=>x(x+1)=20.21
=>x=20
x + 30% . x = \(\frac{1}{3}\)
x ( 1 + \(\frac{3}{10}\)) = \(\frac{1}{3}\)
\(x\left(\frac{10}{10}+\frac{3}{10}\right)=\frac{1}{3}\)
\(x\cdot\frac{13}{10}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{13}{10}=\frac{1}{3}\cdot\frac{10}{13}=\frac{10}{39}\)
#Hoq chắc _ Baccanngon
\(x+30\%x=\frac{1}{3}\)
\(x+\frac{3}{10}x=\frac{1}{3}\)
\(x.1+\frac{3}{10}x=\frac{1}{3}\)
\(x\left(1+\frac{3}{10}\right)=\frac{1}{3}\)
\(x.\frac{13}{10}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{13}{10}\)
\(x=\frac{10}{39}\)
~Học tốt~