Câu hỏi của Hackpro2404

Question

a)Cho a,b,c>0. Chứng minh rằng: $M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$không là số nguyênb)Tính: $A=1^2+2^2+3^2+...+99^2+100^2$

✓ ℍɠŞ_ŦƦùM $₦G ✓ · Accepted Answer

$a.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$1<M<2$\Rightarrow M\notin N$$\Rightarrowđpcm$b.$A=1^2+2^2+...+100^2$$=1+\left(1+1\right)2+\left(1+2\right)3+...+\left(1+99\right)100$$=\left(1+2+3+...+100\right)+\left(1.2+2.3+...+99.100\right)$đặt $N=1.2+2.3+...+99.100$$\Rightarrow3N=1.2.3+2.3.3+3.4.3+...+99.100.3$$=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)$$=1.2.3-1.2.3+2.3.4-2.3.4+...+99.100.101$$=99.100.101\Rightarrow N=\frac{99.100.101}{3}=333300$$\Rightarrow A=5050+333300=338350$ Câu trả lời: $a.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$1<M<2$\Rightarrow M\notin N$$\Rightarrowđpcm$b.$A=1^2+2^2+...+100^2$$=1+\left(1+1\right)2+\left(1+2\right)3+...+\left(1+99\right)100$$=\left(1+2+3+...+100\right)+\left(1.2+2.3+...+99.100\right)$đặt $N=1.2+2.3+...+99.100$$\Rightarrow3N=1.2.3+2.3.3+3.4.3+...+99.100.3$$=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)$$=1.2.3-1.2.3+2.3.4-2.3.4+...+99.100.101$$=99.100.101\Rightarrow N=\frac{99.100.101}{3}=333300$$\Rightarrow A=5050+333300=338350$

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