1.次の等式を証明しなさい。
(1)\(a^3-b^3=(a-b)^3+3ab(a-b)\)
(右辺)\(=(a-b)^3+3ab(a-b)\)
\(\ \ \ \ \ \ \ \ =a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(\ \ \ \ \ \ \ \ =a^3-b^3\)
よって、
\(a^3-b^3=(a-b)^3+3ab(a-b)\)
(2)\((a^2+1)(b^2+1)=(ab+1)^2+(a-b)^2\)
(左辺)\(=(a^2+1)(b^2+1)\)
\(\ \ \ \ \ \ \ \ =a^2b^2+a^2+b^2+1\)
(右辺)\(=(ab+1)^2+(a-b)^2\)
\(\ \ \ \ \ \ \ \ =a^2b^2+2ab+1+a^2-2ab+b^2\)
\(\ \ \ \ \ \ \ \ =a^2b^2+a^2+b^2+1\)
よって、
\((a^2+1)(b^2+1)=(ab+1)^2+(a-b)^2\)
(3)\(x^2+y^2+z^2+xy+yz+zx\)
\(\displaystyle =\frac{1}{2}\{(x+y)^2+(y+z)^2+(z+x)^2\}\)
(右辺)\(\displaystyle =\frac{1}{2}\{(x+y)^2+(y+z)^2+(z+x)^2\}\)
\(\displaystyle =\frac{1}{2}(x^2+2xy+y^2+y^2+2yz+z^2+z^2+2zx+x^2)\)
\(\displaystyle =\frac{1}{2}(2x^2+2y^2+2z^2+2xy+2yz+2zx)\)
\(=x^2+y^2+z^2+xy+yz+zx\)
よって、
\(x^2+y^2+z^2+xy+yz+zx\)
\(\displaystyle =\frac{1}{2}\{(x+y)^2+(y+z)^2+(z+x)^2\}\)
(4)\(a+b+c=0\)のとき、\(a^2+ca=b^2+bc\)
\(c=-a-b\)より、
(左辺)\(=a^2+ca\)
\(\ \ \ \ \ \ \ \ =a^2+a(-a-b)\)
\(\ \ \ \ \ \ \ \ =a^2-a^2-ab)\)
\(\ \ \ \ \ \ \ \ =-ab)\)
(右辺)\(=b^2+bc\)
\(\ \ \ \ \ \ \ \ =b^2+b(-a-b)\)
\(\ \ \ \ \ \ \ \ =b^2-ab-b^2\)
\(\ \ \ \ \ \ \ \ =-ab)\)
よって、
\(a^2+ca=b^2+bc\)
(5)\(a+b+c=0\)のとき、\(ab(a+b)+bc(b+c)+ca(c+a)+3abc=0\)
\(a+b=-c,b+c=-a,c+a=-b\)より、
(左辺)
\(=ab(a+b)+bc(b+c)+ca(c+a)+3abc\)
\(=ab(-c)+bc(-a)+ca(-b)+3abc\)
\(=0\)
よって、
\(ab(a+b)+bc(b+c)+ca(c+a)+3abc=0\)
(6)\(a+b+c=0\)のとき、\(a^3+b^3+c^3=3abc\)
\(c=-a-b\)より、
(左辺)\(=a^3+b^3+c^3\)
\(\ \ \ \ \ \ \ \ =a^3+b^3+(-a-b)^3\)
\(\ \ \ \ \ \ \ \ =a^3+b^3-(a^3+3a^2b+3ab^2+b^3)\)
\(\ \ \ \ \ \ \ \ =-3a^2b-3ab^2\)
(右辺)\(=3abc\)
\(\ \ \ \ \ \ \ \ =3ab(-a-b)\)
\(\ \ \ \ \ \ \ \ =-3a^2b-3ab^2\)
よって、
\(a^3+b^3+c^3=3abc\)
(7)\(a+b+c=0\)のとき、\(a^2-b^2-c^2-2bc=0\)
\(c=-a-b\)より、
(左辺)\(=a^2-b^2-c^2-2bc\)
\(\ \ \ \ \ \ \ \ =a^2-b^2-(-a-b)^2-2b(-a-b)\)
\(\ \ \ \ \ \ \ \ =a^2-b^2-a^2-2ab-b^2+2ab+2b^2\)
\(\ \ \ \ \ \ \ \ =0\)
よって、
\(a^2-b^2-c^2-2bc=0\)
(8)\(\displaystyle \frac{a}{b}=\frac{c}{d}\)のとき、\(\displaystyle \frac{a+c}{b+d}=\frac{2a-3c}{2b-3d}\)
\(\displaystyle \frac{a}{b}=\frac{c}{d}=k\)とおくと、\(a=bk,c=dk\)
(左辺)\(\displaystyle =\frac{a+c}{b+d}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{bk+dk}{b+d}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{k(b+d)}{b+d}\)
\(\ \ \ \ \ \ \ \ =k\)
(右辺)\(\displaystyle =\frac{2a-3c}{2b-3d}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{2bk-3dk}{2b-3d}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{k(2b-3d)}{2b-3d}\)
\(\ \ \ \ \ \ \ \ =k\)
よって、
\(\displaystyle \frac{a+c}{b+d}=\frac{2a-3c}{2b-3d}\)
(9)\(\displaystyle \frac{a}{b}=\frac{c}{d}\)のとき、\(\displaystyle \frac{a^2+c^2}{b^2+d^2}=\frac{a^2}{b^2}\)
\(\displaystyle \frac{a}{b}=\frac{c}{d}=k\)とおくと、\(a=bk,c=dk\)
(左辺)\(\displaystyle =\frac{a^2+c^2}{b^2+d^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{b^2k^2+d^2k^2}{b^2+d^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{k^2(b^2+d^2)}{b^2+d^2}\)
\(\ \ \ \ \ \ \ \ =k^2\)
(右辺)\(\displaystyle =\frac{a^2}{b^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{b^2k^2}{b^2}\)
\(\ \ \ \ \ \ \ \ =k^2\)
よって、
\(\displaystyle \frac{a^2+c^2}{b^2+d^2}=\frac{a^2}{b^2}\)
2.\(\displaystyle \frac{a+b}{3}=\frac{b+c}{4}=\frac{c+a}{5}\neq0\)のとき、\(\displaystyle \frac{ab+bc+ca}{a^2+b^2+c^2}\)の値を求めなさい。
\(\displaystyle \frac{a+b}{3}=\frac{b+c}{4}=\frac{c+a}{5}=k\)とおくと、
\(\left\{\begin{array}{l}a+b=3k \\ b+c=4k \\ c+a=5k\end{array}\right.\)
これを解くと、
\(a=2k,b=k,c=3k\)
よって、
\(\displaystyle \frac{ab+bc+ca}{a^2+b^2+c^2}\)
\(\displaystyle =\frac{2k^2+3k^2+6k^2}{(2k)^2+k^2+(3k)^2}\)
\(\displaystyle =\frac{11k^2}{14k^2}\)
\(k\neq0\)より、
\(\displaystyle =\frac{11}{14}\)
