Toán Lớp 9: Cho P=$\frac{2x+2}{ √x}$ + $\frac{x √x-1}{x- √x}$ – $\frac{x √x+1}{x-+√x}$ (x>0,xkhác 1)
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Toán Lớp 9: Cho P=$\frac{2x+2}{ √x}$ + $\frac{x √x-1}{x- √x}$ – $\frac{x √x+1}{x-+√x}$ (x>0,xkhác 1)
Rút gọn P
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ĐKXĐ: x>0;xne1
=\frac{2x+2}{\sqrt{x}}+\frac{(\sqrt{x})^3-1}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x})^3+1}{\sqrt{x}(\sqrt{x}+1)}
=\frac{2x+2}{\sqrt{x}}+\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}
=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}
=\frac{2x+2+x+\sqrt{x}+1-(x-\sqrt{x}+1)}{\sqrt{x}}
=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}
=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}
Vậy P=\frac{2x+2+2\sqrt{x}}{\sqrt{x}} với x>0;xne1