According to the questionx=3–√+2–√y=3–√−2–√(x3−202–√)−(y3+22–√)=[(3–√+2–√)3−202–√−(3–√−2–√)2−22–√]=33–√+22–√+92–√+63–√−202–√−33–√+22–√+92–√−63–√−22–√=93–√−92–√−93–√+92–√=0
According to the questionx=3–√+2–√y=3–√−2–√(x3−202–√)−(y3+22–√)=[(3–√+2–√)3−202–√−(3–√−2–√)2−22–√]=33–√+22–√+92–√+63–√−202–√−33–√+22–√+92–√−63–√−22–√=93–√−92–√−93–√+92–√=0
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