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We subtitute \(ab+bc+ca=1\) into \(a^2+1\). We have: \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)\)\(=\left(a+b\right)\left(a+c\right)\)
Similarly, we have \(b^2+1=\left(a+b\right)\left(b+c\right)\) and \(c^2+1=\left(a+c\right)\left(b+c\right)\)
From these, we have \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)\(=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Thus, we must have \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)\(=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)
Because both \(a,b,c\) are rational numbers, \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) must be a rational number. Therefore, \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\) is also a rational number.
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)
\(=\sqrt{\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)}\)
\(=\sqrt{\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(b+c\right)+a\left(b+c\right)\right]\left[c\left(c+a\right)+b\left(c+a\right)\right]}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}\)
\(=\sqrt{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)
Do \(a,b,c\) là các số hữu tỉ nên \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) là số hữu tỉ.
\(\Rightarrowđpcm\)
\(a,b\ne0\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{b^2\left(a+b\right)^2+a^2\left(a+b\right)^2+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2b^2+2ab^3+b^4\right)+\left(a^4+2a^3b+a^2b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^4+2a^2b^2+b^4\right)+2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+b^2+ab\right)^2}{a^2b^2\left(a+b\right)^2}}=\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\)
Do \(a,b\) là số hữu tỉ nên \(\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\) cũng là số hữu tỉ.
\(\Rightarrowđpcm\)
Bạn nên viết đề bằng công thức toán để được hỗ trợ tốt hơn. Đọc thế này khó hiểu lắm.
`a)`\(P=A:B\)
\(P=\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\)
\(P=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(P=\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-x^2}{\left(x-1\right)\left(x+1\right)}\)
\(P=-\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)
`b)`\(P=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(2x^2+2\right)=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow4x^2+4=x^2-1\)
\(\Leftrightarrow3x^2=-5\) ( vô lý )
Vậy không có giá trị `x` thỏa mãn `P=1/2`
help^^