[Toán 7] Tìm x

H

hiensau99

lx+\dfrac{1}{101}l+lx+\dfrac{2}{101}l+...+lx+\dfrac{100}{101}l=101x
Thanhs@};-

Ta có $|x+\dfrac{1}{101}|+|x+\dfrac{2}{101}|+...+|x+ \dfrac{100}{101} |\ge 0 \to 101x \ge 0 \to x \ge 0$

$\to x+ \dfrac{1}{101} > 0; \ x+ \dfrac{2}{101} > 0;... ; x+ \dfrac{100}{101} > 0 \to |x+\dfrac{1}{101}| = x+\dfrac{1}{101}; |x+\dfrac{2}{101}|= x+\dfrac{2}{101}; ... ; |x+\dfrac{100}{101}|= x+\dfrac{100}{101}$

Thay vào đẳng thức ta có:

$x+\dfrac{1}{101}+x+\dfrac{2}{101}+...+x+\dfrac{100}{101}=101x$

$\to 100x + \dfrac{(1+100).100:2}{101}= 101x$

$\to x = 50$



 
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