Bài 6( lưu ý: giải theo lớp 6

`bb6.`

Ta có:

`1/(x+y)+1/(y+z)+1/(x+z)=1/100`

`(x+y+z) . (1/(x+y)+1/(y+z)+1/(x+z))=(x+y+z)/100`

`(x+y+z)/(x+y)+(x+y+z)/(y+z)+(x+y+z)/(x+z)=1`

`((x+y)/(x+y)+z/(x+y))+((y+z)/(y+z)+x/(y+z))+((x+z)/(x+z)+y/(x+z))=1`

`(1+z/(x+y))+(1+x/(y+z))+(1+y/(x+z))=1`

`3+(x/(y+z)+y/(x+z)+z/(x+y))=1`

Thay vào `A`, ta được:

`3+A=1`

`A=1-3`

`A=-2`

Vậy `A=-2`

$\color{#FF2E8A}{♡^♡}
\color{#FF3B94}{𝕻}
\color{#FF4FA3}{𝖍}
\color{#FF61AE}{𝖚}
\color{#FF73B6}{𝖔}
\color{#FF85BF}{𝖓}
\color{#FF97C8}{𝖌}
\color{#FFA9D1}{𝖌} \ 
\color{#FFB9D9}{𝕷}
\color{#FFC9E1}{𝖎}
\color{#FFD6E8}{𝖓}
\color{#FFE3EF}{𝖍}
\color{#FFF0F6}{𝖍}
\color{#FF2E8A}{♡^♡}$