tìm x biết (x + 1) + (x + 2) + (x + 3) + (x + 4 + ... + (x + 99) + (x + 100) = 102x
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a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)
Ta có :
\(\frac{x+1}{100}+\frac{x+2}{99}=\frac{x+3}{98}+\frac{x+4}{97}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{100}+1\right)+\left(\frac{x+2}{99}+1\right)=\left(\frac{x+3}{98}+1\right)+\left(\frac{x+4}{97}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}=\frac{x+101}{98}+\frac{x+101}{97}\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}-\frac{x+101}{98}-\frac{x+101}{97}=0\)
\(\Leftrightarrow\)\(\left(x+101\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
Vì \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\)
Nên \(x+101=0\)
\(\Rightarrow\)\(x=-101\)
Vậy \(x=-101\)
Chúc bạn học tốt ~
(X + 1) + (X + 2) + (X + 3) + … + (X + 99) + (X +100) = 5050
(X+X+X+...+X+X) + (1+100) x 50 = 5050
100 x X + 101 x 50 = 5050
100 x X + 5050 = 5050
100 x X = 5050 - 5050
100 x X = 0
X= 0: 100
X= 0
100x+5050=102x
100x+5050=100x+2x
=>2x=5050
x=5050/2
x=2525
(x + 1) + (x + 2) + ... + (x + 100) = 102x
x + x + x + ... + x + 1 + 2 + ... + 100 = 102x
100x + (100 + 1) x 100 : 2 = 102x
102x - 100x = 5050
2x = 5050
x = 2525