Đáp án:
a, x = 91
b, x = -100
c, x = 5
Giải thích các bước giải:
a, $\frac{x-1}{90}$ + $\frac{x-2}{89}$ + $\frac{x-4}{87}$ = 3
⇔ ($\frac{x-1}{90}$ - 1) + ($\frac{x-2}{89}$ - 1) + ($\frac{x-4}{87}$ - 1) = 0
⇔ $\frac{x-91}{90}$ + $\frac{x-91}{89}$ + $\frac{x-91}{87}$ = 0
⇔ (x - 91).($\frac{1}{90}$ + $\frac{1}{89}$ + $\frac{1}{87}$) = 0
mà $\frac{1}{90}$ + $\frac{1}{89}$ + $\frac{1}{87}$ $\neq$ 0
⇒ x - 91 = 0
⇔ x = 91
b, $\frac{x+1}{99}$ + $\frac{x+2}{98}$ + $\frac{x+3}{97}$ + $\frac{x+4}{96}$ + 4 = 0
⇔ ($\frac{x+1}{99}$ + 1) + ($\frac{x+2}{98}$ + 1) + ($\frac{x+3}{97}$ + 1) + ($\frac{x+4}{96}$ + 1) = 0
⇔ $\frac{x+100}{99}$ + $\frac{x+100}{98}$ + $\frac{x+100}{97}$ + $\frac{x+100}{96}$ = 0
⇔ (x + 100).($\frac{1}{99}$ + $\frac{1}{98}$ + $\frac{1}{97}$ + $\frac{1}{96}$) = 0
mà $\frac{1}{99}$ + $\frac{1}{98}$ + $\frac{1}{97}$ + $\frac{1}{96}$ $\neq$ 0
⇒ x + 100 = 0
⇔ x = -100
c, $2^{x+1}$ - $2^{x}$ = 32
⇔ $2^{x}$.(2 - 1) = $2^{5}$
⇔ $2^{x}$ = $2^{5}$
⇔ x = 5
