Question

IF (3√2-√5)÷3√2+√5)=A+B√10,find A and B.

Answer 1

Answer:

a=$\frac{23}{13}$ or 1

b$\sqrt{10}$=$\frac{6\sqrt{10} }{13}$

b=$\frac{6}{13}$

Step-by-step explanation:

• $\frac{3\sqrt{2}-\sqrt{5} }{3\sqrt{2}+\sqrt{5}} =a+b\sqrt{10}$
• $\frac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}+\sqrt{5}} *\frac{3\sqrt{2}-\sqrt{5}}{3\sqrt{2}-\sqrt{5} }=a+b\sqrt{10}$
• $\frac{({2} 3\sqrt{2}-\sqrt{5}) ^{2} }{(3\sqrt{2} )^{2}-(\sqrt{5})^{2} }=a+b\sqrt{10}$

• $\frac{(3\sqrt{2} )^{2}-2*3\sqrt{2}*\sqrt{5}+(\sqrt{5} )^{2}}{18+5} =a+b\sqrt{10}$

• $\frac{18-6\sqrt{10} +5}{13} =a+b\sqrt{10}$
• $\frac{23-6\sqrt{10} }{13} =a+b\sqrt{10}$
• ∵a=$\frac{23}{13}$
• b$\sqrt{10}$=$\frac{6\sqrt{10} }{13}$
• b=$\frac{6}{13}$