a=√(11-6√2),b=√(12-8√2),求1/a+1/b
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a=√(11-6√2),b=√(12-8√2),求1/a+1/b
![a=√(11-6√2),b=√(12-8√2),求1/a+1/b](/uploads/image/z/15067339-43-9.jpg?t=a%3D%E2%88%9A%2811-6%E2%88%9A2%29%2Cb%3D%E2%88%9A%2812-8%E2%88%9A2%29%2C%E6%B1%821%2Fa%2B1%2Fb)
11-6√2
=9-2√18+2
=(3-√2)²
12-8√2
=8-2√32+4
=(2√2-2)²
所以原式=1/(3-√2)+1/(2√2-2)
=(3+√2)/(9-2)+(2√2+2)/(8-4)
=(13+9√2)/14
=9-2√18+2
=(3-√2)²
12-8√2
=8-2√32+4
=(2√2-2)²
所以原式=1/(3-√2)+1/(2√2-2)
=(3+√2)/(9-2)+(2√2+2)/(8-4)
=(13+9√2)/14
a=√(11-6√2),b=√(12-8√2),求1/a+1/b
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