初三数学题，求解先化简再求值
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原式
=[(x²+2x)/(1+2x)]/[(x²+x-2)/(x+1)]
=[x(x+2)/(1+2x)]/[(x+2)(x-1)/(x+1)]
=[x/(1+2x)]/[(x-1)/(x+1)]
=x(x+1)/[(1+2x)(x-1)]
因为x=√2
原式
=√2(√2+1)/[(1+2√2)(√2-1)] 上下同乘(√2+1)
=√2(√2+1)²/(1+2√2)
=√2(2√2+3)/(1+2√2)上下同乘(1-2√2)
=√2(2√2+3)(1-2√2)/(-7)
=√2(-4√2-5)/(-7)
=(8+√2)/7
祝学习进步
=[(x²+2x)/(1+2x)]/[(x²+x-2)/(x+1)]
=[x(x+2)/(1+2x)]/[(x+2)(x-1)/(x+1)]
=[x/(1+2x)]/[(x-1)/(x+1)]
=x(x+1)/[(1+2x)(x-1)]
因为x=√2
原式
=√2(√2+1)/[(1+2√2)(√2-1)] 上下同乘(√2+1)
=√2(√2+1)²/(1+2√2)
=√2(2√2+3)/(1+2√2)上下同乘(1-2√2)
=√2(2√2+3)(1-2√2)/(-7)
=√2(-4√2-5)/(-7)
=(8+√2)/7
祝学习进步
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