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.

ĐÁP ÁN C

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

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

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

Dấu "=" khi: \(a^{100}\left(a-1\right)^2=0;b^{100}\left(b-1\right)^2=0\)

\(\Leftrightarrow a=0;b=0;a=1;b=1;\)a và b là hoán vị của 0;1

\(\Leftrightarrow P\in\left\{0;1;2\right\}\)

Lời giải:

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

\(\Rightarrow \left\{\begin{matrix} a^{100}(a-1)+b^{100}(b-1)=0(1)\\ a^{101}(a-1)+b^{101}(b-1)=0(2)\end{matrix}\right.\)

\(\Rightarrow a^{100}(a-1)^2+b^{100}(b-1)^2=0\) (lấy (2) trừ (1))

Ta thấy: \(a^{100}(a-1)^2\geq 0; b^{100}(b-1)^2\geq 0, \forall a,b\in\mathbb{R}\)

Do đó để tổng của chúng bằng $0$ thì:

\(a^{100}(a-1)^2=b^{100}(b-1)^2=0\)

\(\Rightarrow \left[\begin{matrix} a=0\\ a=1\end{matrix}\right.\) và \(\Rightarrow \left[\begin{matrix} b=0\\ b=1\end{matrix}\right.\)

Thay vào điều kiện ban đầu suy ra \((a,b)=(1,1); (0;0); (1;0); (0;1)\)

Vậy \(P=a^{2015}+b^{2015}\in \left\{0;1;2\right\}\)

a, = 2017/2016.(1+1/2015) = 2017/2016.2016/2015 = 2017/2015

b, = -9/101-9/102-9/101+9/102 = -18/101

k mk nha

a, \(=2017\left(\frac{1}{2016}+\frac{1}{2015.2016}\right)\)

\(=2017\left(\frac{2015+1}{2016}\right)\)

=2017.1=2017

b, = -9/101-9/102-9/101+9/102

=-18/101

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

@@                                         

a: 2015 x 102

= 2015 x ( 100 + 2 )

=2015 x 100 + 2015 x 2

=201500 + 4030

=205530

b: 998 x 45

=(1000 - 2 ) x 45

=1000 x 45 - 2 x 45

=45000 - 90

=44910

a      2015  x  102 = 2015 x (100 + 2) = 2015 x100 + 2015 x 2 = 201500 + 4030= 205530

b      998 x 45 = (1000 - 2) x 45= 45 x 1000 - 45 x 2= 45000- 90 = 45910 

a: 2015 x 102 = 2015 x ( 100 + 2 ) = 2015 x 100 + 2015 x 2 = 201500 + 1030 = 205530

b: 998 x 45 = ( 1000 - 2 ) x 45 = 1000 x 45 - 2 x 45 = 45000 - 90 = 44910