Tìm x: a) x-1/90 + x-2/89 + x-4/87 = 3 b) x+1/99 + x+2/98 + x+3/97 + x+4/96 + 4 = 0 c) 2^x+1 – 2^x = 32 11/08/2021 Bởi Harper Tìm x: a) x-1/90 + x-2/89 + x-4/87 = 3 b) x+1/99 + x+2/98 + x+3/97 + x+4/96 + 4 = 0 c) 2^x+1 – 2^x = 32
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 Bình luận
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