1.若|(X-1)/(2X-3)|+[(3Y+1)/Y+4]^2=0.求代数式:3/2X+1-2/3Y-1.此题答案为2.2.已知A,B,C满足1/A+1/B+1/C不等于0,A(1/B+1/C)+B(1/A+1/C)+C(1/A+1/B)=-3,求A+B+C的值.此题答案为0.

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1.若|(X-1)/(2X-3)|+[(3Y+1)/Y+4]^2=0.求代数式:3/2X+1-2/3Y-1.此题答案为2.2.已知A,B,C满足1/A+1/B+1/C不等于0,A(1/B+1/C)+B

1.若|(X-1)/(2X-3)|+[(3Y+1)/Y+4]^2=0.求代数式:3/2X+1-2/3Y-1.此题答案为2.2.已知A,B,C满足1/A+1/B+1/C不等于0,A(1/B+1/C)+B(1/A+1/C)+C(1/A+1/B)=-3,求A+B+C的值.此题答案为0.
1.若|(X-1)/(2X-3)|+[(3Y+1)/Y+4]^2=0.求代数式:3/2X+1-2/3Y-1.
此题答案为2.
2.已知A,B,C满足1/A+1/B+1/C不等于0,A(1/B+1/C)+B(1/A+1/C)+C(1/A+1/B)=-3,求A+B+C的值.
此题答案为0.

1.若|(X-1)/(2X-3)|+[(3Y+1)/Y+4]^2=0.求代数式:3/2X+1-2/3Y-1.此题答案为2.2.已知A,B,C满足1/A+1/B+1/C不等于0,A(1/B+1/C)+B(1/A+1/C)+C(1/A+1/B)=-3,求A+B+C的值.此题答案为0.
第一题,
(x-1)/(2x-3) = 0 => x=1
(3y+1)/y + 4 = 0 => y=-1/7
代数式可得解
第二题
设A+B+C = X
原代数式可拆为
A/B + A/C + B/A + B/C + C/A + C/B = -3
=> (A+C)/B + (A+B)/C + (B+C)/A = -3
=> (X-B)/B + (X-C)/C + (X-A)/A = -3
=> X/B -1 + X/C -1 + X/A -1 = -3
=> X(1/B + 1/C + 1/A) = 0
=> X = 0 (因为 1/A + 1/B + 1/C 不等于0)

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