Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)

Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)

Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)

\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)

\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)

\(2^{2010}-M=2^{2010}-1\)

=> M = 1

\(a-b=7\Leftrightarrow b=a-7\)

\(\Rightarrow P=\frac{3a-\left(a-7\right)}{2a-7}+\frac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)

\(=\frac{3a-a+7}{2a-7}+\frac{3a-21-a}{2a-14-7}\)

\(=\frac{2a+7}{2a-7}+\frac{2a-21}{2a-21}\)

\(=\frac{2a+7}{2a-7}+1=\frac{2a+7+2a-7}{2a-7}=\frac{4a}{2a-7}\)

Ta có : \(12x^2+20x+1=4\left[3x^2+5x-2\right]+9=4\cdot0+9=9\)

Vậy giá trị biểu thức là 9

Làm rõ ra hộ mình với bạn ơi

\(M=2x^4+3x^2y^2+y^4+y^2\) với \(x^2+y^2=1\)

\(=2x^2.x^2+2x^2y^2+x^2y^2+y^2y^2+y^2\)

\(=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)

\(=2x^2.1+y^2.1+y^2\)

\(=2x^2+y^2+y^2\)

=\(2\left(x^2+y^2\right)\)

\(=2.1=2\)

\(\Rightarrow M=2\)

a/ Ta có : Góc MNy = Góc MNP - Góc PNy = 140-90 = 50 độ
               Góc PNx = Góc MNP - Góc MNx = 140-90 = 50 độ
             => Góc MNy = Góc PNx
b, Ta có : Góc xNy = Góc MNP - Góc MNy - Góc PNx = 140 - 50 - 50 = 40 độ
c, Ta có : Góc MNz = Góc MNy + Góc yNz
              Góc PNz = Góc PNx + Góc xNz
           mà Góc MNy = Góc PNx ( phần a)
               Nz là phân giác của góc xNy => Góc yNz = Góc xNz
          => Góc MNz = Góc PNz
            => Nz là phân giác góc MNP

TL:

\(B=2x^2+y^2-2xy-2x+3\)

    \(=\left(x^2-2xy+y^2\right)+(x^2-2x+1)+2\)

    \(=\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\forall x;y\)

\(D=\left(x+8\right)^4+\left(x+6\right)^4\ge0\forall x\)

Dấu"=" xảy ra<=> \(\hept{\begin{cases}\left(x+8\right)^4=0\\\left(x+6\right)^4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-8\\x=-6\end{cases}}\)

\(x=\sqrt{x}\Rightarrow x^2-x=0\Rightarrow x\left(x-1\right)\)

\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(x-\sqrt{x}=0\)

\(\Rightarrow x=\sqrt{x}\)\(\Rightarrow x^2=x\)( bình phương 2 vế )

\(\Rightarrow x^2-x=x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(x=0\)hoặc \(x=1\)

\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)

\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)

\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=0\)

\(\frac{x+5+2005}{2005}+\frac{x+6+2004}{2004}+\frac{x+7+2003}{2003}=0\)

\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2012}{2003}=0\)

\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

\(x+2010=0\)

\(x=-2010\)