\(\frac{A}{B}=\frac{2010+2011.2012}{2012.2013-2014}=\frac{2010+2011}{2013-2014}=\frac{4021}{-1}=-4021\)

Ta có:

\(a^{2010}+b^{2010}+a^{2012}+b^{2012}\)

\(=\left(a^{2010}+a^{2012}\right)+\left(b^{2010}+b^{2012}\right)\ge2a^{2011}+2b^{2011}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a^{2010}=a^{2012}\\b^{2010}=b^{2012}\end{cases}}\)

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

\(\Rightarrow a^{2013}+b^{2013}=2\)

Vậy \(S=2\)

thank ban nha

ket qua bang 1

\(\frac{A}{B}=\frac{2010+2011\times2012}{2012\times2013-2014}\)

B = 2012 x 2013 - 2014 = 2012 x (2011+2) - 2014 = 2012 x 2011 + 2012 x 2 - 2014 = 2012 x 2011 + 2010 = 2010 + 2011 x 2012

Thay B vào biểu thức tính thương, ta được:

\(\frac{A}{B}=1\)

Đáp số: 1

Nếu mình giúp đc bạn, thì cho mình nhé!

Bài giải

Ta có: 
2010 + 2011 x 2012 /2012 x 2013 – 2014
= ( 2010 + 2011 x 2012)  / (2012 x (2011 + 2)  – 2014)
= ( 2010 + 2011 x 2012)  / (2012 x 2011) + ((2012 x2 ) – 2014)  
= ( 2010 + 2011 x 2012)  / (2012 x 2011) + 2010
= 1/1
= 1

nhin vao de la bik = 1 rui ko can phai lam dai dong vay dau

=1 nha ủng hộ thì mk ha

​

ta có \(a^{2012}+b^{2012}=a^{2013}+b^{2013}\)

\(\Rightarrow a^{2012}-a^{2013}+b^{2012}_{ }-b^{2013}=0\)

\(\Rightarrow a^{2012}\left(1-a\right)+b^{2012}\left(1-b\right)=0\)\(\left(1\right)\)

tương tự \(a^{2013}+b^{2013}=a^{2014}+b^{2014}\)

\(\Leftrightarrow a^{2013}\left(1-a\right)+b^{2013}\left(1-b\right)=0\)\(\left(2\right)\)

trừ (1) cho (2)

ta có \(\left(a^{2012}-a^{2013}\right)\left(1-a\right)\)\(+\left(b^{2012}-b^{2013}\right)\left(1-b\right)=0\)

\(\Leftrightarrow a^{2012}\left(1-a\right)^2+b^{2012}\left(1-b\right)^2=0\)

mà\(a^{2012}\left(1-a\right)^2\ge0;b^{2012}\left(1-b\right)^2\ge0\)

\(\Rightarrow a=1;b=1\)

\(\Rightarrow M=20\times1+11\times1+2013=2044\)

lay cai dau tru cai thu 2

xong lay cai thu 2 tru cai thu 3

xong lay ket qua dau tim dc tru ket qua sau la tim dc a=b=1

roi thay vao tinh M la xong

Ta có: \(a^{2012}+b^{2012}=a^{2013}+b^{2012}=a^{2014}+b^{2014}\)

\(\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\)

\(\Rightarrow a^{2012}+b^{2012}-2\left(a^{2013}+b^{2013}\right)+a^{2014}+b^{2014}=0\)

\(\Leftrightarrow\left(a^{1006}-a^{1007}\right)^2+\left(b^{1006}-b^{1007}\right)=0\)

Từ đó ta có 2 TH

\(\hept{\begin{cases}a^{1006}-a^{1007}=0\\b^{1006}-b^{1007}=0\end{cases}\hept{\begin{cases}a=0;a=1\\b=0;b=1\end{cases}}}\)

Vậy P=20.0+11.0+2013=2013

       P=20.1+11.0+2013=2033

       P=20.0+11.1+2013=2024

\(A=4048142\)

\(B=4048142\)

\(4048142:4048142=1\)

A : B = 1

2010+2011.2012=2010+4046132=4048142

2012.2013-2014=4050156-2014=4048142