Tìm x biết |x+1/2|+|x+1/6|+|x+1/12|+…………+|x+1/110|=11x 04/12/2021 Bởi Anna Tìm x biết |x+1/2|+|x+1/6|+|x+1/12|+…………+|x+1/110|=11x
Ta thấy : $|x+\dfrac{1}{2}| +|x+\dfrac{1}{6}| + …+|x+\dfrac{1}{110}| ≥ 0$ $⇒11x ≥0 ⇒x≥0$ Khi đó : $x+\dfrac{1}{2}+…+x+\dfrac{1}{110} = 11x$ $⇔ 10x + \dfrac{10}{11} =11x$ $⇔x=\dfrac{10}{11}$ ( Thỏa mãn ) Bình luận
Từ bài ta có: $|x+\dfrac{1}{2}|\geqslant0;|x+\dfrac{1}{6}|\geqslant0;….;|x+\dfrac{1}{110}| \geqslant 0$ $⇔ |x+\dfrac{1}{2}|+|x+\dfrac{1}{6}|+…+|x+\dfrac{1}{110}|$ $⇔ 11x \geqslant 0$ $⇔ x \geqslant 0$ $⇔ x+\dfrac{1}{2} \geqslant 0 ; x+\dfrac{1}{6} \geqslant 0;…; x+\dfrac{1}{110} \geqslant 0$ `⇔ (x+1/2)+(x+1/6)+…+(x+1/110)=11x` `⇔ 10x+(1/2+1/6+…+1/110)=11x` `⇔ 10x + (1/(1.2)+1/(2.3)+…+1/(10.11))=11x` `⇔ 10x + (1-1/2+1/2-1/3+…+1/10-1/11)=11x` `⇔ 10x+10/11=11x` `⇔ x=10/11` Vậy `x=10/11` Bình luận
Ta thấy :
$|x+\dfrac{1}{2}| +|x+\dfrac{1}{6}| + …+|x+\dfrac{1}{110}| ≥ 0$
$⇒11x ≥0 ⇒x≥0$
Khi đó :
$x+\dfrac{1}{2}+…+x+\dfrac{1}{110} = 11x$
$⇔ 10x + \dfrac{10}{11} =11x$
$⇔x=\dfrac{10}{11}$ ( Thỏa mãn )
Từ bài ta có:
$|x+\dfrac{1}{2}|\geqslant0;|x+\dfrac{1}{6}|\geqslant0;….;|x+\dfrac{1}{110}| \geqslant 0$
$⇔ |x+\dfrac{1}{2}|+|x+\dfrac{1}{6}|+…+|x+\dfrac{1}{110}|$
$⇔ 11x \geqslant 0$
$⇔ x \geqslant 0$
$⇔ x+\dfrac{1}{2} \geqslant 0 ; x+\dfrac{1}{6} \geqslant 0;…; x+\dfrac{1}{110} \geqslant 0$
`⇔ (x+1/2)+(x+1/6)+…+(x+1/110)=11x`
`⇔ 10x+(1/2+1/6+…+1/110)=11x`
`⇔ 10x + (1/(1.2)+1/(2.3)+…+1/(10.11))=11x`
`⇔ 10x + (1-1/2+1/2-1/3+…+1/10-1/11)=11x`
`⇔ 10x+10/11=11x`
`⇔ x=10/11`
Vậy `x=10/11`