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Lời giải:
\(P=\left[\frac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}+\frac{x}{\sqrt{x}(\sqrt{x}-1)}\right]:\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right].\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}.\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b. Áp dụng BĐT AM-GM
\(M=P\sqrt{x}=\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)
\(=(\sqrt{x}-1)+\frac{1}{\sqrt{x}-1}+2\geq 2\sqrt{(\sqrt{x}-1).\frac{1}{\sqrt{x}-1}}+2=2+2=4\)
Vậy $M_{\min}=4$ khi $\sqrt{x}-1=\frac{1}{\sqrt{x}-1}$
$\Rightarrow \sqrt{x}-1=0$
$\Leftrightarrow x=1$
\(A=\left(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}\right):\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)
\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)
\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)
\(\Rightarrow A=\dfrac{x^2}{x+1}\)
đk : xkhác -1 ; 1
\(A=\left(\dfrac{1+x-1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{x-x+1}{x\left(x-1\right)}\right)=\dfrac{x}{\left(x+1\right)\left(x-1\right)}:\dfrac{1}{x\left(x-1\right)}=\dfrac{x^2}{x+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)
a) Ta có: \(P=\left(\dfrac{3}{x+1}+\dfrac{x-9}{x^2-1}+\dfrac{2}{1-x}\right):\dfrac{x-3}{x^2-1}\)
\(=\left(\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x-9}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x-3}{x^2-1}\)
\(=\dfrac{3x-3+x-9-2x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x-3}\)
\(=\dfrac{2x-14}{x-3}\)
b) Ta có: \(x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Thay x=-3 vào biểu thức \(P=\dfrac{2x-14}{x-3}\), ta được:
\(P=\dfrac{2\cdot\left(-3\right)-14}{-3-3}=\dfrac{-20}{-6}=\dfrac{10}{3}\)
Vậy: Khi \(x^2-9=0\) thì \(P=\dfrac{10}{3}\)
c) Để P nguyên thì \(2x-14⋮x-3\)
\(\Leftrightarrow2x-6-8⋮x-3\)
mà \(2x-6⋮x-3\)
nên \(-8⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(-8\right)\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(\Leftrightarrow x\in\left\{4;2;5;1;7;-1;11;-5\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;7;11;-5\right\}\)
Vậy: Để P nguyên thì \(x\in\left\{4;2;5;7;11;-5\right\}\)
Với `x \ne -5,x \ne -1` có:
`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`
`A=[x^2-3x-4]/[(x+5)(x+1)]`
`A=[(x-4)(x+1)]/[(x+5)(x+1)]`
`A=[x-4]/[x+5]`
\(=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+x+5x+5}-\dfrac{1}{x+1}\\ =\dfrac{x+2}{x+5}+\dfrac{-5x-1}{\left(x^2+x\right)+\left(5x+5\right)}-\dfrac{1}{x+1}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{x\left(x+1\right)+5\left(x+1\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{\left(x+1\right)\left(x+5\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+2x+x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+x-4x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x^2+x\right)-\left(4x+4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x\left(x+1\right)-4\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x-4}{x+5}\)
a: \(=\dfrac{x^3-x^2+x+3\left(x^2-1\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+4+3x^2-3}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+2x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^2+x+1}{x^2-x+1}\)
b: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=>A>0 với mọi x<>-1
\(=\dfrac{-3\left(x-2\right)-2\left(x+2\right)+4x}{x^2-4}\)
\(=\dfrac{-3x+6-2x-4+4x}{x^2-4}\)
\(=\dfrac{-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{1}{x+2}\left(x\ne2;x\ne-2\right)\)
\(\dfrac{-3}{x+2}-\dfrac{2}{x-2}+\dfrac{4x}{x^2-4}\left(x\ne\pm2\right)\)
\(=\dfrac{-3\left(x-2\right)-2\left(x+2\right)+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-3x+6-2x-4+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{1}{x+2}\)
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
Với x ≠ 0; x ≠ 1, ta có:
(1 - x²)/[x(x - 1)]
= -(x - 1)(x + 1)/[x(x - 1)]
= -(x + 1)/x