(x+1)+(x+2)+........+(x+100) = 15050

100x + 100.(100+1):2 = 15050

100x + 5050 = 15050

100x = 15050 - 5050 = 10000

x = 10000 : 100 = 100 

( x + 1) + ( x + 2 ) + ( x + 3 ) + ........+ ( x + 100 ) = 15050

100x + 100 . ( 100 +1 ) : 2 = 15050

100x = 15050 -5050

100 x = 10 000 

x= 10 000 : 100

x= 100

\(\frac{x}{98}+\frac{x-1}{99}+\frac{x-2}{100}+\frac{1-3}{101}=-4\)

<=>  \(\frac{x}{98}+1+\frac{x-1}{99}+1+\frac{x-2}{100}+1+\frac{x-3}{101}+1=0\)

<=>  \(\frac{x+98}{98}+\frac{x+98}{99}+\frac{x+98}{100}+\frac{x+98}{101}=0\)

<=>  \(\left(x+98\right)\left(\frac{1}{98}+\frac{1}{99}+\frac{1}{100}+\frac{1}{101}\right)=0\)

<=>  \(x+98=0\)  (do 1/98 + 1/99 + 1/100 + 1/101 khác 0)

<=> \(x=-98\)

Vậy...

\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
     \(-12x+60+21-7x=5\)
     \(-12x-7x+81=5\)
     \(-19x=5-81\)
     \(-19x=-76\)
      \(x=-76:\left(-19\right)\)
      \(x=4\)
\(Vậyx=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
     \(30x+60-6x-30-24x=100\)
     \(30x-6x-24x+60-30=100\)
     \(0x+30=100\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn đề bài.
\(c,-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
     \(-5x-1-\frac{1}{2}x-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
      \(-5x-\frac{1}{2}x-1-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
      \(-\frac{11}{2}x-\frac{2}{3}=\frac{3}{2}x-\frac{5}{6}\)
       \(-\frac{2}{3}+\frac{5}{6}=\frac{3}{2}x+\frac{11}{2}x\)
        \(-\frac{4}{6}+\frac{5}{6}=\frac{14}{2}x\)
         \(\frac{1}{6}=7x\)
          \(x=\frac{1}{6}:7\)

         \(x=\frac{1}{6}.\frac{1}{7}\)
         \(x=\frac{1}{42}\)
\(Vậyx=\frac{1}{42}\)
\(d,-3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)

      \(-3x+\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
      \(-3x-5x+\frac{3}{2}-3=-x+\frac{1}{5}\)
      \(-8x+\frac{3}{2}-\frac{6}{2}=-x+\frac{1}{5}\)
       \(-8x-\frac{3}{2}=-x+\frac{1}{5}\)
        \(-\frac{3}{2}-\frac{1}{5}=-x+8x\)
        \(\frac{15}{10}-\frac{2}{10}=7x\)

         \(7x=\frac{13}{10}\)
          \(x=\frac{13}{10}:7\)
          \(x=\frac{13}{10}.\frac{1}{7}\)
          \(x=\frac{13}{70}\)
\(Vậyx=\frac{13}{70}\)

 

Bài 2 : 

a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = 2 

b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)

Dấu ''='' xảy ra khi x = -1 

 Bài 1 : 

a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)

c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

a, 5^{2x - 3} - 2.5² = 5².3

=> 5^{2x - 3} - 50 = 75

=> 5^{2x - 3} = 125

=> 5^{2x - 3} = 5³

=> 2x - 3 = 3

=> 2x = 0

=> x = 0

vậy_

b, (x + 1) + (x + 2) + ....+ (x + 100) = 5750

=> x + 1 + x + 2 + ... + x + 100 = 5750

=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750

=> 100x + 5050 = 5750

=> 100x = 700

=> x = 7

vậy_

a,2x+53=135

2x=135-53

2x=82

x=82:2

x=41

bạn viết khó hỉu quá nên mk giúp bạn dc câu a thui 

k hộ mk với

Tách 2 vế ra bn

A.-12.(x-5)+7.(3-x)=5

=>-12.(x-5)=5 hay 3-x=5

cảm ơn nhé