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Question

Rut gon da thuc sau: X=$\sqrt{a^2+1+\left(1-\frac{1}{a+1}\right)^2}+\frac{a}{a+1}$voi a>0

alibaba nguyễn · Accepted Answer

$X=\sqrt{a^2+1+\left(1-\frac{1}{a+1}\right)^2}+\frac{a}{a+1}$$=\sqrt{a^2+1+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\sqrt{\frac{\left(a^2+1\right)\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1$Câu trả lời: $X=\sqrt{a^2+1+\left(1-\frac{1}{a+1}\right)^2}+\frac{a}{a+1}$$=\sqrt{a^2+1+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\sqrt{\frac{\left(a^2+1\right)\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}$$=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1$

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