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catlt01012004

2 năm trước

Loga Sinh Học lớp 12

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linh2111

Đáp án:

`x=10;y=20/3;z=100/21`

Trình bày lời giải:

`c)`
Ta có:
`2x=3y=>x/3=y/2=>x/21=y/14(1)`
`5y=7z=>y/7=z/5=>y/14=z/10(2)`
Từ `(1)` và `(2)`
`=>x/21=y/14=z/10`
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
`x/21=y/14=z/10=(3x+5z-7y)/(3.21+5.14-7.10)=30/63=10/21`
Do đó:
`{(x/21=10/21=>x=10),(y/14=10/21=>y=14 . 10/21=20/3),(z/10=10/21=>z=10. 10/21=100/21):}`
Vậy `x=10;y=20/3;z=100/21`

Vote (0) Phản hồi (0) 2 năm trước

Tranngocyen160

`2x=3y\to x/3=y/2\to x/21=y/14(1)`

`5y=7z\to y/7=z/5\to y/14=z/10(2)`

Từ `(1)` và `(2)\to x/21=y/14=z/10`

`↔ (3x)/63=(7y)/98=(5z)/50`

mà `3x+5z-7y=30`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`(3x)/63=(7y)/98=(5z)/50=(3x-7y+5z)/(63-98+50)=30/15=2`

`\to x/21=2\to x=2.21=42`

`\to y/14=2\to y=2.14=28`

`\to z/10=2\to z=2.10=20`

Vậy `(x;y;z)=(42;28;20)`

Vote (0) Phản hồi (0) 2 năm trước

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