tìm x biết lx+1/2l+lx+1/6l+lx+1/12l+….+lx+1/110l=11x 07/12/2021 Bởi Piper tìm x biết lx+1/2l+lx+1/6l+lx+1/12l+….+lx+1/110l=11x
`|x+1/2|+|x+1/6|+…+|x+1/110|=11x` VÌ `|x+1/2|≥0 ; |x+1/6|≥0 ;…;|x+1/110|≥0` Nên `|x+1/2|+|x+1/6|+…+|x+1/110|≥0` Suy ra `11x≥0` `⇒x≥0` Ta có: `x+1/2+x+1/6+…+x+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+(1-1/11)=11x` `⇒x=10/11` Bình luận
Đáp án: `|x+1/2|+|x+1/6|+|x+1/12|+….+|x+1/110|=11x` `(1)` Vì `VT>=0 ∀x` `=> VP>=0 ∀x` `=> 11x>=0 ∀x` `=> x >=0 ∀x` `=> x+1/2>0 ∀x, x+1/6 >0 ∀x, x+1/12 >0 ∀x, …. , x+1/(110) >0 ∀x` `(1)` `=> x+1/2+x+1/6+….+x+1/110=11x` `=> (x+x+…+x)+(1/2+1/6+…+1/110)=11x` `=> (x+x+…+x)+(1/(1.2)+1/(2.3)+…+1/(10.11))=11x` `=> 10x+(1/1-1/11)=11x` `=> 10x+10/11=11x` `=> 10/11=11x-10x` `=> 10/11=x` Vậy `x=10/11` < Bài này cô đã chữa, bạn yên tâm tham khảo ^-^ > Bình luận
`|x+1/2|+|x+1/6|+…+|x+1/110|=11x`
VÌ `|x+1/2|≥0 ; |x+1/6|≥0 ;…;|x+1/110|≥0`
Nên `|x+1/2|+|x+1/6|+…+|x+1/110|≥0`
Suy ra `11x≥0`
`⇒x≥0`
Ta có: `x+1/2+x+1/6+…+x+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+(1-1/11)=11x`
`⇒x=10/11`
Đáp án:
`|x+1/2|+|x+1/6|+|x+1/12|+….+|x+1/110|=11x` `(1)`
Vì `VT>=0 ∀x`
`=> VP>=0 ∀x`
`=> 11x>=0 ∀x`
`=> x >=0 ∀x`
`=> x+1/2>0 ∀x, x+1/6 >0 ∀x, x+1/12 >0 ∀x, …. , x+1/(110) >0 ∀x`
`(1)` `=> x+1/2+x+1/6+….+x+1/110=11x`
`=> (x+x+…+x)+(1/2+1/6+…+1/110)=11x`
`=> (x+x+…+x)+(1/(1.2)+1/(2.3)+…+1/(10.11))=11x`
`=> 10x+(1/1-1/11)=11x`
`=> 10x+10/11=11x`
`=> 10/11=11x-10x`
`=> 10/11=x`
Vậy `x=10/11`
< Bài này cô đã chữa, bạn yên tâm tham khảo ^-^ >