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Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{2x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x^2-1}{5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+2x+x+1+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+3x+9-x^2+2x-1}{5}\)
\(=\dfrac{x^2+5x+8}{5}\)
Ta có: \(x^2+5x+8\)
\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)
Ta có: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\)
\(\Leftrightarrow x^2+5x+8>0\forall x\)
\(\Leftrightarrow\dfrac{x^2+5x+8}{5}>0\forall x\) thỏa mãn ĐKXĐ(đpcm)
\(a,A=\dfrac{x+1+2-2x+5-x}{\left(1-x\right)\left(x+1\right)}\cdot\dfrac{\left(1-x\right)\left(x+1\right)}{2x-1}\left(x\ne1;x\ne-1;x\ne\dfrac{1}{2}\right)\\ A=\dfrac{8-2x}{2x-1}\\ b,A>0\Leftrightarrow\dfrac{8-2x}{2x-1}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-2x>0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}8-2x< 0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x< 4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 4\)
1: Sai
2: Sai
3: Đúng
4: Đúng
Lời giải:
1. Đúng
2. Sai
3. Đúng
4. Đúng