Question

1.cho a, b, c thỏa mãn \(a^3+b^3+c^3=3abc\)

tính gtbt: \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

2. tìm nghiệm nguyên dương của phương trình \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{9}{xyz}=1\)

Answer 1

\(a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(\Rightarrow P=\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}=\frac{\left(-c\right)\left(-b\right)\left(-a\right)}{abc}=-1\)

TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Answer 2

b/ \(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}+9.xyz=1\Leftrightarrow x+y+z+9=xyz\)

Không mất tính tổng quát, giả sử \(x\le y\le z\)

Nếu \(z< 3\Rightarrow VP\le8< 9< VT\Rightarrow ptvn\) \(\Rightarrow z\ge3\)

\(\Rightarrow x+y+z+9\le3z+9\le3\left(z+3\right)\le6z\Rightarrow xyz\le6z\)

\(\Rightarrow xy\le6\Rightarrow\left(x;y\right)=\left(1;1\right);\left(1;2\right);\left(1;3\right);\left(1;4\right);\left(1;5\right);\left(1;6\right);\left(2;3\right)\)

- Nếu \(\left(x;y\right)=\left(1;1\right)\Rightarrow z+11=z\left(l\right)\)

- Nếu \(\left(x;y\right)=\left(1;2\right)\Rightarrow z+12=2z\Rightarrow z=12\)

- Nếu \(\left(x;y\right)=\left(1;3\right)\Rightarrow z+13=3z\left(l\right)\)

- Nếu ....