Đặt `a/b=c/d=k` `=> a=bk;c=dk` Thay vào ta có : `[7a^2+3ab]/[11a^2-8b^2]=[7(bk)^2+3bkb]/[11(bk)^2-8b^2]=[b^2(7k^2+3k)]/[b^2(11k^2-8)]=[7k^2+3k]/[11k^2-8]` `(1)` `[7c^2+3cd]/[11c^2-8d^2]=[7(dk)^2+3dkd]/[11(dk)^2-8d^2]=[d^2(7k^2+3k)]/[d^2(11k^2-8)]=[7k^2+3k]/[11k^2-8]` `(2)` Từ `(1),(2)=>(7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)` (Đpcm)
Đáp án:
Giải thích các bước giải:
`a/b=c/d`
`ĐK:b,d \ne 0`
`+)c=0`
`=>a/b =0`
`=>a=0`
`=>(7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)=0`
`+)c \ne 0` ,`a/b=c/d`
`=>a/c=b/d`
Đặt `a/c=b/d=k (k \ne 0)`
`=>a=ck,b=dk`
`=>(7a^2+3ab)/(7c^2+3cd)`
`=[7(ck)^2+3.ck.dk]/(7c^2+3cd)`
`=(7c^2 k^2+3cd.k^2)/(7c^2+3cd)`
`=(k^2(7c^2+3cd))/(7c^2+3cd)`
`=k^2`
`=>(11a^2-8b^2)/(11c^2-8d^2)`
`=[11(ck)^2-8(dk)^2]/(11c^2-8d^2)`
`=(11c^2 k^2+8d^2 k^2)/(11c^2-8d^2)`
`=(k^2(11c^2-8d^2))/(11c^2-8d^2)`
`=k^2`
`=>(7a^2+3ab)/(7c^2+3cd)=(11a^2-8b^2)/(11c^2-8d^2) (=k^2)`
`=>(7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)`
Đáp án + Giải thích các bước giải:
Đặt `a/b=c/d=k`
`=> a=bk;c=dk`
Thay vào ta có :
`[7a^2+3ab]/[11a^2-8b^2]=[7(bk)^2+3bkb]/[11(bk)^2-8b^2]=[b^2(7k^2+3k)]/[b^2(11k^2-8)]=[7k^2+3k]/[11k^2-8]` `(1)`
`[7c^2+3cd]/[11c^2-8d^2]=[7(dk)^2+3dkd]/[11(dk)^2-8d^2]=[d^2(7k^2+3k)]/[d^2(11k^2-8)]=[7k^2+3k]/[11k^2-8]` `(2)`
Từ `(1),(2)=>(7a^2+3ab)/(11a^2-8b^2)=(7c^2+3cd)/(11c^2-8d^2)` (Đpcm)