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a) Sửa đề: \(\left(ax+by+cx\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2\)
= a2x2 + b2y2 + c2x2 + 2axby + 2bycz + 2axcz + b2x2 - 2bxay + a2y2 + c2y2 - 2cybz + b2z2 + a2z2 - 2azcx + c2x2
= a2x2 + b2y2 + c2x2 + b2x2 + a2y2 + c2y2 + b2z2 + a2z2 + c2x2
= a2(x2+y2+z2) + b2(x2+y2+z2) + c2(x2+y2+z2)
= (a2+b2+c2)(x2+y2+z2) (đpcm)
b) Đặt x = b; y = c; z = a, ta có:
\(\left(ay+bz+cx\right)^2+\left(az-by\right)^2+\left(bx-cz\right)^2+\left(cy-ax\right)^2\)
= a2y2 + b2z2 + c2x2 + 2aybz + 2bzcx + 2aycx + a2z2 - 2azby + b2y2 + b2x2 - 2bxcz + c2z2 + c2y2 - 2cyax + a2x2
= a2y2 + b2z2 + c2x2 + a2z2 + b2y2 + b2x2 + c2z2 + c2y2 + a2x2
= (a2+b2+c2)(x2+y2+z2)
Thay b = x, c = y, a = z, ta có:
(a2+b2+c2)(x2+y2+z2) = (a2+b2+c2)2 (đpcm)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b: \(=\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)-abc\)
\(=\left(a+b\right)\left(ab+bc+ca\right)+abc+c\left(bc+ca\right)-abc\)
\(=\left(a+b\right)\cdot\left(ab+ac+bc\right)+c^2\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\cdot\left(b+c\right)\left(a+c\right)\)
c:\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-b^4+6a^3b-6ab^3+8ab^3-8a^3b\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)+6ab\left(a-b\right)\left(a+b\right)+8ab\left(b-a\right)\left(b+a\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2+6ab-8ab\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)