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DK: x>1
\(P=\dfrac{1}{x}\left(\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2+\left(\sqrt{x+1}-\sqrt{x-1}\right)^2}{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\right)\)
\(P=\dfrac{1}{x}\left(\dfrac{x+1+x-1+2\sqrt{x^2-1}+x+1+x-1-2\sqrt{x^2-1}}{x+1-x+1}\right)\)\(P=\dfrac{1}{x}\cdot\dfrac{4x}{2}\)
\(P=2\)
\(P=\dfrac{1}{x}\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}+\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\right)=\dfrac{1}{x}\left[\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2+\left(\sqrt{x+1}-\sqrt{x-1}\right)^2}{\sqrt{\left(x+1\right)}^2-\sqrt{\left(x-1\right)^2}}\right]=\dfrac{1}{x}\left(\dfrac{x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}+x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}+x-1}{x+1-x+1}\right)=\dfrac{1}{x}\cdot\dfrac{4x}{2}=\dfrac{1}{x}\cdot2x=2\)
=> Giá trị của biểu thức P không phụ thuộc vào biến
\(A=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+y}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(x-y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x+y}\cdot\dfrac{x+\sqrt{xy}-\sqrt{xy}+y}{x-y}\)
\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x-y}\)
\(=\dfrac{\sqrt{xy}+y-x-\sqrt{xy}-y}{x-y}=\dfrac{-x}{x-y}\)
\(C=\dfrac{1}{\left(\dfrac{x+2\sqrt{xy}+y-x-y}{\left(\sqrt{x+y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)^2}-\dfrac{x+y}{2\sqrt{xy}}-\dfrac{\left(x+y\right)^2}{4xy}\)
\(=\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}{4xy}-\dfrac{\left(x+y\right)^2}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{\left(x+y\right)\left(x+y+2\sqrt{xy}\right)-\left(x+y\right)^2}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{2\sqrt{xy}\left(x+y\right)}{4xy}-\dfrac{x+y}{2\sqrt{xy}}\)
\(=\dfrac{x+y-x-y}{2\sqrt{xy}}=0\)
\(=\dfrac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+12x+22\sqrt{x}+13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
biểu thức này có phụ thuộc vào biến nha bạn
Ta có: \(A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\left(x-2\sqrt{xy}+y\right)}{x-y}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
=1
Hướng dẫn trả lời:
ĐKXĐ: 0 < x ≠ 1.
Đặt √x = a (a > 0 và a ≠ 1)
Ta có:
(2+√xx+2√x+1−√x−2x−1).x√x+x−√x−1√x=[2+aa2+2a+1−a−2a2−1].a3+a2−a−1a=[(2+a)(a−1)−(a−2)(a+1)(a+1)(a2−1)].(a+1)(a2−1)a=2a(a+1)(a2−1).(a+1)(a2−1)a=2