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NeTakaya
3 years ago
13

{\sqrt{x} +1}{x-\sqrt{x} +1}" alt="\frac{\sqrt{x} +1}{x-\sqrt{x} +1}" align="absmiddle" class="latex-formula"> Với x≥0. Tìm GTLN
Mathematics
1 answer:
zubka84 [21]3 years ago
8 0

Answer:

Step-by-step explanation:

\frac{\sqrt{x}+1 }{x-\sqrt{x}+1 }=\frac{\sqrt{x}+1}{(x+1)-\sqrt{x}}\\\\=\frac{(\sqrt{x}+1)([x+1]+\sqrt{x})}{([x+1]-\sqrt{x})+([x+1)+\sqrt{x}])}\\\\=\frac{\sqrt{x}*x+\sqrt{x}*1+\sqrt{x}*\sqrt{x}+1*x+1*1+1*\sqrt{x}}{(x+1)^{2}-(\sqrt{x})^{2}}\\\\\\=\frac{x\sqrt{x}+\sqrt{x}+x+1+x+\sqrt{x}}{x^{2}+2x+1-x}\\\\=\frac{x\sqrt{x}+2\sqrt{x}+2x+1}{x^{2}+x+1}\\\\

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