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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 \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a. ĐK: a, b, c khác 0.
\(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}=1\)
\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)
\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)
\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)
\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).
b) Không mất tính tổng quả. G/s: a = b + c
Khi đó ta có:
\(\frac{a^2+b^2-c^2}{2ab}=\frac{\left(b+c\right)^2+b^2-c^2}{2\left(b+c\right)b}=1\)
\(\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b+c\right)^2}{2bc}=-1\)
\(\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+\left(b+c\right)^2-b^2}{2\left(b+c\right)c}=1\)
=> Điều phải chứng minh.
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{ab+ac+bc}{abc}=\frac{1}{abc}\)
\(\Rightarrow ab+ac+bc=1\)
Ta có :
\(1+a^2=ab+ac+bc+a^2=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)
\(1+b^2=ab+ac+bc+b^2=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\)
\(1+c^2=ab+ac+bc+c^2=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là bình phương của 1 số hữu tỉ (ĐPCM)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow ab+bc+ca=1\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)\(=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)
\(=\left(a+c\right)\left(b+a\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{abc}\left(QĐ\right)\Leftrightarrow ac+bc+ab=1\)
\(\Rightarrow1+a^2=bc+ab+ac+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+c\right)\left(a+b\right)\)
Tương tự: \(1+b^2=\left(a+b\right)\left(b+c\right)\); \(1+c^2=\left(a+c\right)\left(b+c\right)\)
Nhân vế với vế ta được: \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)
mà \(\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)là số chính phương => đpcm
Đặt \(\frac{a}{b^2}=x;\frac{b}{c^2}=y;\frac{c}{a^2}=z\) thì \(\frac{b^2}{a}=\frac{1}{x};\frac{a^2}{c}=\frac{1}{y};\frac{c^2}{b}=\frac{1}{z}\)
\(\Rightarrow xyz=\frac{a}{b^2}\cdot\frac{b}{c^2}\cdot\frac{c}{a^2}=1\)
Ta có: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=xy+yz+zx\)
Lại có: \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=xyz-xy-yz-zx+x+y+z-1=1-x-y-z+x+y+z-1=0\)(vì xyz=1, xy+yz+zx=x+y+z)
=>x-1=0 hoặc y-1=0 hoặc z-1=0
=>x=1 hoặc y=1 hoặc z=1
=>a/b2=1 hoặc b/c2=1 hoặc c/a2=1
=>a=b2 hoặc b=c2 hoặc c=a2 (ĐPCM)
Cách khác
Ta có: \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{b^2}{a}+\frac{a^2}{c}+\frac{c^2}{b}\)
<=>\(a^2b^2c^2\left(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\right)=abc\left(\frac{b^2}{a}+\frac{a^2}{c}+\frac{c^2}{b}\right)\) (a2b2c2=abc=1)
<=>\(\frac{a^3b^2c^2}{b^2}+\frac{a^2b^3c^2}{c^2}+\frac{a^2b^2c^3}{a^2}=\frac{ab^3c}{a}+\frac{a^3bc}{c}+\frac{abc^3}{b}\)
<=>\(a^3c^2+b^3a^2+c^3b^2=b^3c+a^3b+c^3a\)
<=>\(a^3c^2+b^3a^2+c^3b^2-b^3c-a^3b-c^3a-a^2b^2c^2+abc=0\) (a2b2c2=abc=1)
<=>\(\left(a^3c^2-a^2b^2c^2\right)+\left(b^3a^2-a^3b\right)+\left(c^3b^2-c^3a\right)+\left(abc-b^3c\right)=0\)
<=>\(-a^2c^2\left(b^2-a\right)+a^2b\left(b^2-a\right)+c^3\left(b^2-a\right)-bc\left(b^2-a\right)=0\)
<=>\(\left(b^2-a\right)\left(-a^2c^2+a^2b+c^3-bc\right)=0\)
<=>\(\left(b^2-a\right)\left[c^2\left(c-a^2\right)-b\left(c-a^2\right)\right]=0\)
<=>\(\left(b^2-a\right)\left(c^2-b\right)\left(c-a^2\right)=0\)
Đến đây dễ rồi