3/ Ta có:

\(x+y+z=0\)

\(\Rightarrow x^2=\left(y+z\right)^2;y^2=\left(z+x\right)^2;z^2=\left(x+y\right)^2\)

\(a+b+c=0\)

\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Ta có:

\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)

\(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\)

\(=-ax^2-by^2-cz^2\)

\(\Leftrightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Leftrightarrow ax^2+by^2+cz^2=0\)

1/ Đặt \(a-b=x,b-c=y,c-z=z\)

\(\Rightarrow x+y+z=0\)

Ta có:

\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

2/ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)

\(\Leftrightarrow ab+bc+ca=1\)

Ta có:

\(M=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)

\(=\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)\)

\(=\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)\left(c+a\right)\left(c+b\right)\)

\(=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

ta có \(\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2=\)\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}+2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\right)\)

= \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\)= A²

vậy A = \(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)là một số hữu tỉ

Câu hỏi của Phạm Quang Dương - Toán lớp 9 - Học toán với OnlineMath

Đặt \(a-b=x;b-c=y;c-a=z\)

\(\Rightarrow x+y+z=a-b+b-c+c-a=0\)

Lúc đó: \(B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Mà \(x+y+z=0\Rightarrow2\left(x+y+z\right)=0\Rightarrow\frac{2\left(x+y+z\right)}{xyz}=0\)

\(\Rightarrow B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)

\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{xz}+\frac{2}{xy}\)

\(=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

Ta có: \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\)

\(=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2-2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\right)\)

\(=\left(\frac{1}{\left(a-b\right)}+\frac{1}{\left(b-c\right)}+\frac{1}{c-a}\right)^2-2\left(\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)

\(=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)

=> \(A=\sqrt{\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}}=\sqrt{\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2}\)

\(=\left|\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right|\)

Vì a,b,c là các số hữu tỉ => \(\left|\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right|\)là một số hữu tỉ

=> A là một số hữu tỉ

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_ Hok tốt _

1/ Đặt

\(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z,xyz=1\)thì ta có

\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow xy+yz+zx=x+y+z\)

\(\Leftrightarrow xyz-xy-yz-zx+x+y+z-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)

\(\Leftrightarrow x=1;y=1;z=1\)

\(\Rightarrow\frac{a}{b^2}=1;\frac{b}{c^2}=1;\frac{c}{a^2}=1\)

\(\Leftrightarrow a=b^2;b=c^2;c=a^2\)

2/ Đặt

\(ab=x,bc=y,ca=z\) cần tính

\(P=\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\left(1+\frac{y}{x}\right)\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)

Xét \(x+y+z=0\)

\(\Rightarrow P=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-1\)

Xét \(x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

mà nè đề bài câu 1 là abc=1 chứ có phải xyz=1 đâu