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\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)
\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)
\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)
\(C=c\left(a-b\right)\left(c^2+ab-ac-bc\right)\)
\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)
\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)
\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
\(\frac{b+c+d}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}=\frac{\left(a+b+c+d-x\right)+\left(x-a\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}\)\(=\frac{\left(a+b+c+d-x\right)}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}+\frac{1}{\left(b-a\right)\left(c-a\right)\left(d-a\right)}\)
Áp dụng hoán vị vòng \(b\rightarrow c\rightarrow d\rightarrow a\rightarrow b\) vào VT , ta được :
\(\left(a+b+c+d-x\right)\)[\(\frac{1}{\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(a-x\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)\left(b-d\right)\left(b-x\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)\left(c-d\right)\left(c-x\right)}\)\(+\frac{1}{\left(d-a\right)\left(d-b\right)\left(d-c\right)\left(d-x\right)}\).
Quy đồng mẫu thức và tính toán biểu thức trong [ ] ta được :
\(\frac{-1}{\left(x-a\right)\left(x-b\right)\left(x-c\right)\left(x-d\right)}\)
Vậy ...............
Ta có:
\(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)
\(=bc\left(a+d\right)\left[\left(b-a\right)+\left(a-c\right)\right]-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\)\(\left(a-b\right)\)
\(=bc\left(a+d\right)\left(a-b\right)+bc\left(a+d\right)\left(a-c\right)-ac\left(b+d\right)\left(a-c\right)\)\(+ab\left(c+d\right)\left(a-b\right)\)
\(=b\left(a-b\right)\left[a\left(c+d\right)-c\left(a+d\right)\right]+c\left(a-c\right)\left[b\left(a+d\right)-a\left(b+d\right)\right]\)
\(=b\left(a-b\right).d\left(a-c\right)+c\left(a-c\right).d\left(b-a\right)\)
\(=d\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
Sao ở đa thức C có x vậy, đề cho x hay sao thế?
Đroi á bn đề bài này có x nha!