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Cho các số a, b, c, x, y, z thoả mãn $\left \{ {{x= by + cz} \atop {y= cz + ax}} \right.$ và z= ax + by. Biết a, b, c khác -1. Tính A= $\frac{1}{1+a}
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Đáp án:
Ta có :
`x + y + z`
`= (by + cz) + (cz + ax) + (ax + by)`
`= (by + by) + (cz + cz) + (ax + ax)`
`= 2by + 2cz + 2ax`
`= 2(ax + by + cz)`
Mặt khác :
`x = by + cz`
`=> x + ax = ax + by + cz`
`=> x(a + 1) = ax + by + cz`
`=> 1/(a + 1) = x/(ax + by + cz)`
`y = cz + ax`
`=> y + by = ax + by + cz`
`=> y(1 + b) = ax + by + cz`
`=> 1/(1 + b) = y/(ax + by + cz)`
`z = ax + by`
`=> z + cz = ax + by + cz`
`=> z(1 + c) = ax + by + cz`
`=> 1/(1 + c) = z/(ax + by + cz)`
Do đó :
`A = 1/(1 + a) + 1/(1 + b) + 1/(1 + c)`
`= x/(ax + by + cz) + y/(ax + by + cz) + z/(ax + by + cz)`
`= (x + y + z)/(ax + by + cz)`
Thay `x + y + z = 2(ax + by + cz)`
`=> A = [2(ax + by + cz)]/(ax + by + cz) = 2`
Giải thích các bước giải: