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cuncon2395

cuncon2395 said:


bài 76: cho biểu thức
[TEX]Q=\frac{a}{\sqrt{a^2-b^2}}-(1+\frac{a}{\sqrt{a^2-b^2}}):\frac{b}{a-\sqrt{a^2-b^2}}[/TEX]

a, Rút g0n Q

b, Xác định giá trị Q khi a=3b
Bấm để xem đầy đủ nội dung ...

[TEX]Q=\frac{a}{\sqrt{a^2-b^2}}-(1+\frac{a}{\sqrt{a^2-b^2}}):\frac{b}{a-\sqrt{a^2-b^2}}[/TEX]
[TEX]=\frac{a}{\sqrt{a^2-b^2}}-\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}[/TEX]
[TEX]= \frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}[/TEX]
[TEX]=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{a^2-b^2}}[/TEX]
[TEX]=\frac{(a-b)\sqrt{a^2-b^2}}{(a-b)(a+b)}=\frac{\sqrt{a^2-b^2}}{a+b}[/TEX]

b, thay a=3b vào
[TEX]Q=\frac{\sqrt{9b^2-b^2}}{3b+b}=\frac{2b\sqrt{2}}{4b}=\frac{\sqrt{2}}{2}[/TEX]
 
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