Toán 9 Rút gọn $A=\sqrt{1+\dfrac1{a^2}+\dfrac1{(a+1)^2}}$

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[Tư Âm Diệp Ẩn]

a) [tex]A^2=1+\frac{1}{a^2}+\frac{1}{(a+1)^2}=\frac{a^2(a+1)^2+(a+1)^2+a^2}{a^2(a+1)^2}=\frac{a^2(a^2+2a+1+1)+(a+1)^2}{a^2(a+1)^2}\\ =\frac{a^4+2a^2(a+1)+(a+1)^2}{a^2(a+1)^2}=\frac{(a^2+a+1)^2}{a^2(a+1)^2}\\ \Rightarrow A=\frac{a^2+a+1}{a(a+1)}(a>0)[/tex]
b) Từ câu a, ta có:
[tex]\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}=\frac{a^2+a+1}{a(a+1)}=1+\frac{1}{a(a+1)}=1+\frac{1}{a}-\frac{1}{a+1}[/tex]
[tex]\Rightarrow B=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\\ B=(1+\frac{1}{1}-\frac{1}{2})+(1+\frac{1}{2}-\frac{1}{3})+....+(1+\frac{1}{99}-\frac{1}{100})\\ B=99+(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100})\\ B=100-\frac{1}{100}=99.99[/tex]

 

[Blue Plus]

View attachment 78179

$1+\dfrac1{a^2}+\dfrac1{(a+1)^2}\\=1+\dfrac{1}{a^2}+\dfrac{1}{[-(a+1)]^2}\\=1+\dfrac{1}{a^2}+\dfrac{1}{[-(a+1)]^2}+2\left ( \dfrac{1+a-(a+1)}{1.a.[-(a+1)]} \right )(\text{do }1+a-(a+1)=0)\\=1+\dfrac{1}{a^2}+\dfrac{1}{[-(a+1)]^2}+2\left ( \dfrac{-1}{a(a+1)}+\dfrac{-1}{a+1}+\dfrac{1}{a} \right )\\=\left ( 1+\dfrac{1}{a}-\dfrac{1}{a+1} \right )^2\\\Rightarrow A=\sqrt{\left(1+\dfrac1a-\dfrac1{a+1}\right)^2}=\left|1+\dfrac1a-\dfrac1{a+1}\right|=1+\dfrac1a-\dfrac1{a+1}(\text{do }a>0)$
$B=\left ( 1+\dfrac{1}{1}-\dfrac{1}{2} \right )+\left ( 1+\dfrac{1}{2}-\dfrac{1}{3} \right )+...+\left ( 1+\dfrac{1}{99}-\dfrac{1}{100} \right )\\...$