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đê yêu cầu CM  \(a^{2007}+b^{2007}+a^{2009}+b^{2009}\) chia hết cho 5

\(x^2-x-1=0\)

Ta có \(\Delta=b^2-4ac=\left(-1\right)^2-4.1.\left(-1\right)=1+4=5>0\); \(\sqrt{\Delta}=\sqrt{5}\)

Phuông trình có 2 nghiệm phân biệt 

\(a=x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{1+\sqrt{5}}{2}\)

\(b=x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{1-\sqrt{5}}{2}\)

Ta có \(a^{2007}+b^{2007}+a^{2009}+b^{2009}\)

\(\Leftrightarrow a^{2007}.\left(1+a^2\right)+b^{2007}.\left(1+b^2\right)\)

\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(1+\left(\frac{1+\sqrt{5}}{2}\right)^2\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(1+\left(\frac{1-\sqrt{5}}{2}\right)^2\right)\)

\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(1+\frac{3+\sqrt{5}}{2}\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(1+\frac{3-\sqrt{5}}{2}\right)\)

\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(\frac{5+\sqrt{5}}{2}\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(\frac{5-\sqrt{5}}{2}\right)\)

\(\Leftrightarrow\sqrt{5}.\left(\frac{1+\sqrt{5}}{2}\right)^{2008}+\sqrt{5}.\left(\frac{1-\sqrt{5}}{2}\right)^{2008}\)

\(\Leftrightarrow\sqrt{5}.\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2008}+\left(\frac{1-\sqrt{5}}{2}\right)^{2008}\right]⋮5\)  (ĐPCM)

Nhớ k cho mình nhé 

Có VT = n³- n + 2017n = n(n²-1) + 2017n = n(n-1)(n+1) + 2017n

Vì (n-1)n(n+1) là tích 3 số tự nhiên liên tiếp nên \(\text{(n-1)n(n+1)}⋮3\)

mà \(2007⋮3\Rightarrow2007n⋮3\)

suy ra \(VT⋮3\) (1)

Lại có 2008 : 3 dư 1=> 2008²⁰⁰⁷:3 dư 1

=> 2008²⁰⁰⁷+1 : 3 dư 2

suy ra VP : 3 dư 2 (2)

Từ (1) và (2) => không tìm được n TM đkđb (đpcm)

Ta có:\(\left(\sqrt[]{x^2+2007}+x^{ }\right)\left(\sqrt{x^2+2007}-x\right)\left(\sqrt{y^2+2007}+y\right)\left(\sqrt{y^2+2007}-y\right)=2007\left(\sqrt{x^2+2007}-x\right)\left(\sqrt{y^2+2007}-y\right)\)

\(\Rightarrow2007^2=2007\left(\sqrt{x^2+2007}-x\right)\left(\sqrt{y^2+2007}-y\right)\)

\(\Rightarrow\left(\sqrt{x^2+2007}-x\right)\left(\sqrt{y^2+2007}-y\right)=2007\)

\(\Rightarrow xy-x\sqrt{y^2+2007}-y\sqrt{x^2+2007}+\sqrt{\left(x^2+2007\right)\left(y^2+2007\right)}=2007\)(1)

và \(\left(\sqrt[]{x^2+2007}+x^{ }\right)\left(\sqrt{y^2+2007}+y\right)=xy+x\sqrt{y^2+2007}+y\sqrt{x^2+2007}+\sqrt{\left(x^2+2007\right)\left(y^2+2007\right)}=2007\)(2)

cộng (1) và (2)

\(\Rightarrow xy+\sqrt{\left(x^2+2007\right)\left(y^2+2007\right)}=2007\)

\(\Leftrightarrow\sqrt{\left(x^2+2007\right)\left(y^2+2007\right)}=2007-xy\)

\(\Rightarrow x^2y^2+2007\left(x^2+y^2\right)+2007^2=2007^2-2.2007xy+x^2y^2\)

\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow M=0\)

thank you Bertram Đức Anh

Ta có : \(\left(a+\sqrt{a^2+2007}\right)\left(-a+\sqrt{a^2+2007}\right)=2007\)

\(\left(b+\sqrt{b^2+2007}\right)\left(-b+\sqrt{b^2+2007}\right)=2007\)

Nhân từng vế với nhau , ta có :

\(\left(a+\sqrt{a^2+2007}\right)\left(b+\sqrt{b^2+2007}\right)\left(-b+\sqrt{b^2+2007}\right)\left(-a+\sqrt{a^2+2007}\right)=2007\)

⇔ \(2007\left(-b+\sqrt{b^2+2007}\right)\left(-a+\sqrt{a^2+2007}\right)=2007^2\)

⇔ \(ab-b\sqrt{a^2+2007}-a\sqrt{b^2+2007}+\sqrt{\left(a^2+2007\right)\left(b^2+2007\right)}=2007\left(1\right)\)

Ta có : \(\left(a+\sqrt{a^2+2007}\right)\left(b+\sqrt{b^2+2007}\right)=2007\)

⇔ \(ab+a\sqrt{b^2+2007}+b\sqrt{a^2+2007}+\sqrt{\left(a^2+2007\right)\left(b^2+2007\right)}=2007\left(2\right)\)

Cộng từng vế của ( 1 ; 2 ) , ta có :

\(ab+\sqrt{\left(a^2+2007\right)\left(b^2+2007\right)}=2007\)

⇔ \(\sqrt{\left(a^2+2007\right)\left(b^2+2007\right)}=2007-ab\)

⇔ \(a^2b^2+2007a^2+2007b^2+2007^2=2007^2-2.2007ab+a^2b^2\)

⇔ \(2007a^2+2007b^2=-2.2007ab\)

⇔ \(a^2+2ab+b=0\)

⇔ \(\left(a+b\right)^2=0\)

⇔ \(S=a+b=0\)