Toán Chứng minh rằng nếu 1/a+1/b+1/c=2 và a+b+c=abc thì ta có 1/a^2+1/b^2+1/c^2=2 30/06/2021 By Mackenzie Chứng minh rằng nếu 1/a+1/b+1/c=2 và a+b+c=abc thì ta có 1/a^2+1/b^2+1/c^2=2
Ta có: `1/a+1/b+1/c=2` `⇔(1/a+1/b+1/c)^2=2^2` `⇔1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=4` `⇔1/a^2+1/b^2+1/c^2+2(1/(ab)+1/(bc)+1/(ca))=4` `⇔1/a^2+1/b^2+1/c^2+2(c/(abc)+a/(abc)+b/(abc))=4` `⇔1/a^2+1/b^2+1/c^2+2.(a+b+c)/(abc)=4` Mà `a+b+c=abc` `\to 1/a^2+1/b^2+1/c^2+2.(abc)/(abc)=4` `⇔1/a^2+1/b^2+1/c^2+2=4` `⇔1/a^2+1/b^2+1/c^2=4-2` `⇔1/a^2+1/b^2+1/c^2=2` Vậy `1/a^2+1/b^2+1/c^2=2(đpcm)` Trả lời
Đáp án + Giải thích các bước giải: `1/a^2+1/b^2+1/c^2=2` `=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=4` `=>1/a^2+1/b^2+1/c^2+2(1/(ab)+1/(bc)+1/(ca))=4` `=>1/a^2+1/b^2+1/c^2+2((a+b+c)/(abc))=4` `=>1/a^2+1/b^2+1/c^2+2((abc)/(abc))=4` `=>1/a^2+1/b^2+1/c^2+2.1=4` `=>1/a^2+1/b^2+1/c^2=2` Vậy `1/a+1/b+1/c=2` và `a+b+c=abc` thì `1/a^2+1/b^2+1/c^2=2` Trả lời
Ta có:
`1/a+1/b+1/c=2`
`⇔(1/a+1/b+1/c)^2=2^2`
`⇔1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=4`
`⇔1/a^2+1/b^2+1/c^2+2(1/(ab)+1/(bc)+1/(ca))=4`
`⇔1/a^2+1/b^2+1/c^2+2(c/(abc)+a/(abc)+b/(abc))=4`
`⇔1/a^2+1/b^2+1/c^2+2.(a+b+c)/(abc)=4`
Mà `a+b+c=abc`
`\to 1/a^2+1/b^2+1/c^2+2.(abc)/(abc)=4`
`⇔1/a^2+1/b^2+1/c^2+2=4`
`⇔1/a^2+1/b^2+1/c^2=4-2`
`⇔1/a^2+1/b^2+1/c^2=2`
Vậy `1/a^2+1/b^2+1/c^2=2(đpcm)`
Đáp án + Giải thích các bước giải:
`1/a^2+1/b^2+1/c^2=2`
`=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=4`
`=>1/a^2+1/b^2+1/c^2+2(1/(ab)+1/(bc)+1/(ca))=4`
`=>1/a^2+1/b^2+1/c^2+2((a+b+c)/(abc))=4`
`=>1/a^2+1/b^2+1/c^2+2((abc)/(abc))=4`
`=>1/a^2+1/b^2+1/c^2+2.1=4`
`=>1/a^2+1/b^2+1/c^2=2`
Vậy `1/a+1/b+1/c=2` và `a+b+c=abc` thì `1/a^2+1/b^2+1/c^2=2`