Bài ni t mần cho phát chán nó  rồi:))

Ta có:\(x^{2012}+y^{2012}=\left(x^{2011}+y^{2011}\right)\left(a+b\right)-ab\left(a^{2010}+b^{2010}\right)\left(1\right)\)

Mặt khác:\(x^{100}+y^{100}=x^{101}+y^{101}=x^{102}+y^{102}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow1=x+y-xy\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow1+y^{2010}=1+y^{2011}=1+y^{2012}\Rightarrow y=1\\y=1\Rightarrow x^{2010}+1=x^{2011}+1=x^{2012}+1\Rightarrow x=1\end{cases}}\)vì \(x;y\) là các số dương

Thay vào ta được:\(A=1^{2020}+1^{2020}=2\)

Làm lại nha.sơ suất quá:((

Ta có:

\(x^{2012}+y^{2012}=\left(x^{2011}+y^{2011}\right)\left(x+y\right)-xy\left(x^{2010}+y^{201}\right)\left(1\right)\)

Mặt khác:\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\left(2\right)\)

Từ (1);(2) suy ra:

\(x^{2010}+y^{2010}=\left(x^{2010}+y^{2010}\right)\left(x+y\right)-xy\left(x^{2010}+y^{2010}\right)\)

\(=\left(x^{2010}+y^{2010}\right)\left(x+y-xy\right)\)

\(\Rightarrow x+y-xy=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow1+y^{2010}=1+y^{2011}=1+y^{2012}\Rightarrow y=1\\y=1\Rightarrow1+x^{2010}=1+x^{2011}=1+x^{2012}\Rightarrow x=1\end{cases}}\)

Thay vào ta được \(A=3\)

Vậy A=3

Lời giải:

a)

$x^3-6x^2-9x+14=0$

$\Leftrightarrow x^3-x^2-5x^2+5x-14x+14=0$

$\Leftrightarrow x^2(x-1)-5x(x-1)-14(x-1)=0$

$\Leftrightarrow (x-1)(x^2-5x-14)=0$

$\Leftrightarrow (x-1)(x^2-7x+2x-14)=0$

$\Leftrightarrow (x-1)[x(x-7)+2(x-7)]=0$

$\Leftrightarrow (x-1)(x+2)(x-7)=0$

$\Rightarrow x=1; x=-2$ hoặc $x=7$

b)

Bạn tham khảo tại đây:

Câu hỏi của Lương Đức Hưng - Toán lớp 8 | Học trực tuyến

a) \(x^3-6x^2-9x+14=0\)

\(\Leftrightarrow x^3-8x^2+2x^2+7x-16x+14=0\)

\(\Leftrightarrow\left(x^3-8x^2+7x\right)+\left(2x^2-16x+14\right)=0\)

\(\Leftrightarrow x\left(x^2-8x+7\right)+2\left(x^2-8x+7\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+7\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-7x-x+7\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-7\right)-\left(x-7\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow x\in\left\{-2;1;7\right\}\)

lấy cả 2 vế trừ đi 3

\(\frac{x-2010-2011}{2009}+\frac{x-2009-2011}{2010}+\frac{x-2009-2010}{2011}=3\)

\(\Leftrightarrow\left(\frac{x-2010-2011}{2009}-1\right)+\left(\frac{x-2009-2011}{2010}-1\right)+\left(\frac{x-2009-2010}{2011}-1\right)=0\)

\(\Leftrightarrow\frac{x-6030}{2009}+\frac{x-6030}{2010}+\frac{x-6030}{2011}=0\)

\(\Leftrightarrow\left(x-6030\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)

\(\Leftrightarrow x-6030=0\)(vì \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}>0\))

\(\Leftrightarrow x=6030\)

Vậy ................