Помощ за уравнение

[t0bl3ro_bg]

Може ли да ми помогнете да решим тази задача, че не се получава отговор [tex]\sqrt{a^2-1}[/tex]

http://a6.sphotos.ak.fbcdn.net/hphotos- ... 3250_n.jpg

[Xixibg]

[tex]A=1-\frac{\sqrt{a+1}-(a+1)\sqrt{a-1}}{\sqrt{a-1}-\sqrt{a+1}}:\frac{\sqrt{a+1}\sqrt{a^2-1}}{(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}}[/tex]

[tex]A=1-\frac{(\sqrt{a+1}-(a+1)\sqrt{a-1})(\sqrt{a-1}+\sqrt{a+1})}{(\sqrt{a-1}-\sqrt{a+1})(\sqrt{a-1}+\sqrt{a+1})}.\frac{(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}}{(a+1)\sqrt{a-1}}[/tex]

[tex]A=1+\frac{(\sqrt{a+1}\sqrt{a-1}-(a+1)(a-1)+(a+1)-(a+1)\sqrt{a-1}\sqrt{a+1}}{2}.\frac{(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}}{(a+1)\sqrt{a-1}}[/tex]

[tex]A=1+\frac{(a+1)(a-1)\sqrt{a-1}-(a+1)(a-1)(a-1)\sqrt{a+1}-(a+1)(a+1)\sqrt{a-1}+(a+1)(a-1)(a+1)\sqrt{a+1}}{2(a+1)\sqrt{a-1}}[/tex]

[tex]A=1+\frac{(a-1)\sqrt{a-1}-(a-1)(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}+(a-1)(a+1)\sqrt{a+1}}{2\sqrt{a-1}}[/tex]

[tex]A=1+\frac{-2\sqrt{a-1}+2(a-1)\sqrt{a+1}}{2\sqrt{a-1}}[/tex]


[tex]A=1+\frac{-\sqrt{a-1}\sqrt{a-1}+(a-1)\sqrt{a+1}\sqrt{a-1}}{\sqrt{a-1}\sqrt{a-1}}[/tex]

[tex]A=1-1+\sqrt{a+1}\sqrt{a-1}[/tex]

[tex]A=\sqrt{a^2-1}[/tex]

[t0bl3ro_bg]

asddsa.jpg (2.25 MiB) Прегледано 218 пъти

Мерси, а би ли помогнал и за тази 47 B)

[Xixibg]

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[tex]A=1+\frac{(\sqrt{a+1}\sqrt{a-1}-(a+1)(a-1)+(a+1)-(a+1)\sqrt{a-1}\sqrt{a+1}}{2}.\frac{(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}}{(a+1)\sqrt{a-1}}[/tex]

[tex]A=1+\frac{(a+1)(a-1)\sqrt{a-1}-(a+1)(a-1)(a-1)\sqrt{a+1}+\cancel{(a+1)(a-1)\sqrt{a+1}}-\cancel{(a+1)(a-1)(a+1)\sqrt{a-1}}}{2(a+1)\sqrt{a-1}}[/tex]
[tex]+\frac{-\cancel{(a+1)(a-1)\sqrt{a+1}}+\cancel{(a+1)(a-1)(a+1)\sqrt{a-1}}-(a+1)(a+1)\sqrt{a-1}+(a+1)(a-1)(a+1)\sqrt{a+1}}{2(a+1)\sqrt{a-1}}[/tex]

[tex]A=1+\frac{(a+1)(a-1)\sqrt{a-1}-(a+1)(a-1)(a-1)\sqrt{a+1}-(a+1)(a+1)\sqrt{a-1}+(a+1)(a-1)(a+1)\sqrt{a+1}}{2(a+1)\sqrt{a-1}}[/tex]

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Ето го междинният ред между 3-ти и 4-ти.....