先化简再求值[(x+2)/(2x²-4x)][x-2+(x-2)/(8x)] 其中x=根号2-1
先化简再求值[(x+2)/(2x²-4x)][x-2+(x-2)/(8x)]其中x=根号2-1...
先化简再求值[(x+2)/(2x²-4x)][x-2+(x-2)/(8x)] 其中x=根号2-1
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1个回答
2017-10-28
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[(x+2)/(2x²-4x)][x-2+(x-2)/(8x)]
=(x+2)(x-2)[1+1/(8x)]/[2x(x-2)]
=(x+2)(x-2)(8x+1)/[16x²(x-2)]
=(x+2)(8x+1)/(16x²)
=(√2-1+2)(8√2-8+1)/[16(√2-1)²]
=(√2+1)(8√2-7)(√2+1)²/16
=(16+√2-7)(3+2√2)/16
=(9+√2)(3+2√2)/16
=(31+21√2)/16
=(x+2)(x-2)[1+1/(8x)]/[2x(x-2)]
=(x+2)(x-2)(8x+1)/[16x²(x-2)]
=(x+2)(8x+1)/(16x²)
=(√2-1+2)(8√2-8+1)/[16(√2-1)²]
=(√2+1)(8√2-7)(√2+1)²/16
=(16+√2-7)(3+2√2)/16
=(9+√2)(3+2√2)/16
=(31+21√2)/16
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