Ta có : a¹⁰² + b¹⁰² = (a¹⁰¹ + b¹⁰¹)(a + b) - ab(a¹⁰⁰ + b¹⁰⁰)

Mà a¹⁰⁰ + b¹⁰⁰ = a¹⁰¹ + b¹⁰¹ = a¹⁰² + b¹⁰². 

Do đó : a + b - ab = 1

=> a + b - ab - 1 = 0

<=> (a - ab) + (b - 1) = 0

<=> a(1 - b) - (1 - b) = 0

=> (a - 1)(1 - b) = 0

\(\Leftrightarrow\orbr{\begin{cases}a-1=0\\1-b=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)

Nên a = 1 thì b = 1 

Vậy  P = a²⁰⁰⁴ + b²⁰⁰⁴ = 1²⁰⁰⁴ + 1²⁰⁰⁴ = 1 + 1 = 2

I have a crazy idea tham khảo nhé:

Vì: a¹⁰⁰ + b¹⁰⁰; a¹⁰¹ + b¹⁰¹; a¹⁰² + b¹⁰² đều = nhau nên a chỉ = 1 => a²⁰⁰⁴ + b²⁰⁰⁴ = 1²⁰⁰⁴ + 1²⁰⁰⁴ = 1 + 1 = 2

Vậy:

khó thế :'(

\(a^{101}+b^{101}=a^{100}+b^{100}\Leftrightarrow a^{101}-a^{100}+b^{101}-b^{100}=0\)

\(\Leftrightarrow a^{100}\left(a-1\right)+b^{100}\left(b-1\right)=0\left(1\right)\)

\(a^{102}+b^{102}=a^{101}+b^{101}\Leftrightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)=0\left(2\right)\)

Trừ vế cho vế của (2) và (1):

\(\left(a-1\right)\left(a^{101}-a^{100}\right)+\left(b-1\right)\left(b^{101}-b^{100}\right)=0\)

\(\Leftrightarrow\left(a-1\right)a^{100}\left(a-1\right)+\left(b-1\right)b^{100}\left(b-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2.a^{100}+\left(b-1\right)^2b^{100}=0\)

Do \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\a^{100}\ge0\\\left(b-1\right)^2\ge0\\b^{100}\ge0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)^2a^{100}+\left(b-1\right)^2b^{100}\ge0\)

Dấu "=" xảy ra khi và chỉ khi: \(\left(a;b\right)=\left(1;1\right);\left(1;0\right);\left(0;1\right);\left(0;0\right)\)

- Nếu \(\left(a;b\right)=\left(1;1\right)\Rightarrow S=1+1=2\)

- Nếu \(\left[{}\begin{matrix}\left(a;b\right)=\left(1;0\right)\\\left(a;b\right)=\left(0;1\right)\end{matrix}\right.\) \(\Rightarrow S+1+0=1\)

- Nếu \(\left(a;b\right)=\left(0;0\right)\) \(\Rightarrow S=0\)

Câu a) 

Em tham khảo link: Câu hỏi của I have a crazy idea - Toán lớp 6 - Học toán với OnlineMath

Ta có bài toán

P_n-P_n-1=(n-1)P_n-1

Chứng minh

Ta có    P_n-P_n-1=n!-(n-1)!

                         =n(n-1)!-(n-1)!

                         =(n-1)(n-1)!=(n-1)P_n-1

=>P_n-P_n-1=(n-1)P_n-1

Từ kết quả trên ta có

P₂-P₁=(2-1)P₁

P₃-P₂=(3-1)P₂

...............

P_n=P_n-1=(n-1)P_n-1

-----------------------------

P_n-P₁=P₁+2P₂+3P₃+.........+(n-1)P₁

=>1+1.P₁+2P₂+3P₃+...+n.P_n=P_n+1

Theo đề ra, ta có:

\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Leftrightarrow\left(a^{100}+b^{100}\right).\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)+a^{202}+b^{202}=a^{202}+b^{202}+2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)=2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2-2ab\right)=0\)

\(\Leftrightarrow a=b=0\)

\(\Rightarrow a^{100}+b^{100}=a^{101}+b^{101}\)

\(\Rightarrow a^{100}=a^{101}\)

\(\Leftrightarrow a^{100}.\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow A=a^{2015}+b^{2015}=1+1=2\).

\(Từ:\) \(a^{100}+b^{100}=a^{101}+b^{101}\)

\(\Leftrightarrow a^{100}\left(a-1\right)+b^{100}\left(b-1\right)=0\left(1\right)\)

\(và\) \(a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Leftrightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)=0 \left(2\right)\)

\(Từ\left(1\right)\) \(và\) \(\left(2\right)\)

\(\Rightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)-a^{100}\left(a-1\right)-b^{100}\left(b-1\right)=0\)

\(\Leftrightarrow a^{100}\left(a-1\right)^2+b^{100}\left(b-1\right)^2\)

\(Do\) \(a,b>0\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Rightarrow A=1+1=2\)

em không chắc cho lắm ạ

hình như là bằng 2 thì phải.

Đặt M=a²⁰⁰⁷+b²⁰⁰⁷

Do \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)(1)

\(\Rightarrow\left(a^{101}+b^{101}\right)^2=\left(a^{100}+b^{100}\right)\left(a^{102}+b^{102}\right)\)

\(\Leftrightarrow a^{202}+b^{202}+2.a^{101}.b^{101}=a^{202}+a^{100}.b^{102}+a^{102}.b^{100}+b^{202}\)

\(\Leftrightarrow2.a^{101}.b^{101}=a^{100}.b^{100}\left(a^2+b^2\right)\)

\(\Leftrightarrow a^{100}.b^{100}\left(a^2-2ab+b^2\right)=0\)

\(\Leftrightarrow a^{100}.b^{100}\left(a-b\right)^2=0\)

Do a,b > 0 => (a-b)²=0 <=> a=b

Thay a=b vào (1) ta được

\(2.a^{100}=2.a^{101}=2.a^{102}\)

\(\Leftrightarrow a^{100}=a^{101}\)

\(\Leftrightarrow a^{100}\left(a-1\right)=0\)

Do a>0 nên a=1 =>b=1

Vậy M=1²⁰¹⁷+1²⁰¹⁷=2

đây nhé hơi dài Xem câu hỏi

๖ۣۜᔕᑌᖇᐯIᐯ.IO [TEᗩᗰ ᗩᔕᗰOᗷIᒪE ᗪOᖇᗩYᗩK]

mình mới học lớp 5

tk nhé@@@@@@@@@@@@@@@@

hihi

LOL

Liên MIh hay s mà LOL?

kih dân net

Ta có đẳng thức: \(a^{102}+b^{102}=\left(a^{101}+b^{101}\right)\left(a+b\right)-ab\left(a^{100}+b^{100}\right)\) với mọi số a,b

Kết hợp với: \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Rightarrow1=\left(a+b\right)-ab\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\Rightarrow1+b^{100}=1+b^{101}=1+b^{102}\Rightarrow b=1\\b=1\Rightarrow1+a^{100}=1+a^{101}=1+a^{102}\Rightarrow a=1\end{matrix}\right.\)

Do đó: \(P=a^{2014}+b^{2014}=1^{2004}+1^{2005}=2\)

(1) => 2x²+2y²+2z²-2xy-2yz-2xz=0

<=> (x²-2xy+y²) + (x^2-2xz+z²) + (y²-2yz+z²)=0

<=> (z-y)² + (x-z)² + (y-z)² = 0

<=> x=y=z

(2) => x²⁰⁰²  + x²⁰⁰²  + x²⁰⁰² = 3²⁰⁰³

<=> 3x²⁰⁰² = 3²⁰⁰³

x=y=z=3

Từ \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}=2\left(a^{101}+b^{101}\right)\)

\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}-2\left(a^{101}+b^{101}\right)=0\)

\(\Rightarrow\left(a^{102}-2a^{101}+a^{100}\right)+\left(b^{102}-2b^{101}+b^{100}\right)=0\)

\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2=0\left(1\right)\)

Vif \(\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2\ge0\forall a\\\left(b^{51}-b^{50}\right)^2\ge0\forall b\end{cases}}\)

\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2\ge0\forall a,b\left(2\right)\)

Tứ (1) và (2) :

\(\Rightarrow\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2=0\\\left(b^{51}-b^{50}\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a^{51}-a^{50}=0\\b^{51}-b^{50}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a^{51}=a^{50}\\b^{51}=b^{50}\end{cases}}\)

Vì a,b là các số thực dương nên \(a=b=1\)

\(\Rightarrow P=a^{2007}+b^{2007}=1^{2007}+1^{2007}=1+1=2\)

Vậy \(P=2\)