Giải giùm minh mấy bài toán này vs ạ 1. If a is divisible by 4 then remainder of [2×(a+1)^2:16]. 2. If a^3+b^3=637 and a+b=13. Find the value of(a-)^2

0 bình luận về “Giải giùm minh mấy bài toán này vs ạ 1. If a is divisible by 4 then remainder of [2×(a+1)^2:16]. 2. If a^3+b^3=637 and a+b=13. Find the value of(a-)^2”

1. 1.Cause $a$ is a divisible by $4$

   $⇒a=4k(k∈N)$

   $⇒2(a+1)^2=2(4k+1)^2=2.16k^2+16k+2$

   So the remainder is $2$ because $2.16k^2;16k\vdots 16$

   2.$a^3+b^3=637$

   $(a+b)(a^2-ab+b^2)=637$

   $13.(a^2-ab+b^2)=637$

   $a^2-ab+b^2=49$

   $a^2-2ab+b^2=49-ab$

   $(a-b)^2=49-ab$

   $a^3+b^3=637$

   $(a+b)^3-3ab(a+b)=637$

   $13^3-3ab.13=637$

   $ab=40$

   $⇒(a-b)^2=49-40=9$

   3.$M=x^2+y^2+2xy+2x+2y+11$

   $=(x+y)^2+2(x+y)+1+10$

   $=(x+y+1)^2+10≥10$

   The equation is $x+y+1=0$

   $⇔x+y=-1$

   Vậy $Min_{M}=10$ với $x+y=-1$

    

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