タイムアタック 目標2分展開せよ(1)(x+1)(x+3)(2)(x−5)(x+3)(3)(2x+3)(x−2)(4)(x+2y−1)(x+2y+1)因数分解せよ(1)x2−3x+2(2)x2−2xy−35y2(3)2x2−3x−2(4)x2+9y2−6xy−1
解答(1)(x+1)(x+3)=x2+x+3x+3=x2+4x+3答え x2+4x+3(2)(x−5)(x+3)=x2−5x+3x−15=x2−2x−15答え x2−2x−15(3)(2x+3)(x−2)=2x2+3x−4x−6=2x2−x−6答え 2x2−x−6(4)(x+2y−1)(x+2y+1)={(x+2y)−1}{(x+2y)+1}=(x+2y)2−12=x2+4xy+4y2−1(x2+4y2+4x−1)答え x2+4xy+4y2−1(x2+4y2+4x−1)因数分解(1)x2−3x+2=(x−1)(x−2)答え (x−1)(x−2)(2)x2−2xy−35y2=(x−7y)(x+5y)答え (x−7y)(x+5y)(3)2x2−3x−2=(2x+1)(x−2)答え (2x+1)(x−2)(4)x2+9y2−6xy−1=(x2−6xy+9y2)−1=(x−3y)2−1=(x−3y+1)(x−3y−1)答え (x−3y+1)(x−3y−1)
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