1 1 1 6
——————— + —————- +…..+ —————– = ————-
x^2 + x x^2+3x+2 x^2+19x+90 x+10
giúp mình với
Đáp án: $x=\dfrac53$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+…+\dfrac{1}{x^2+19x+90}=\dfrac{6}{x+10}$
$\to \dfrac{1}{x(x+1)}+\dfrac{1}{(x+1)(x+2)}+…+\dfrac{1}{(x+9)(x+10)}=\dfrac{6}{x+10}$
$\to \dfrac{(x+1)-x}{x(x+1)}+\dfrac{(x+2)-(x+1)}{(x+1)(x+2)}+…+\dfrac{(x+10)-(x+9)}{(x+9)(x+10)}=\dfrac{6}{x+10}$
$\to\dfrac1x-\dfrac1{x+1}+\dfrac1{x+1}-\dfrac1{x+2}+…+\dfrac1{x+9}-\dfrac1{x+10}=\dfrac6{x+10}$
$\to \dfrac1{x}-\dfrac1{x+10}=\dfrac6{x+10}$
$\to \dfrac1{x}=\dfrac7{x+10}$
$\to x+10=7x$
$\to 6x=10$
$\to x=\dfrac53$