a) \(\orbr{x=0}\)

btvn mà lên mạng thế

a)\(2007x\left(x-\frac{2006}{7}\right)=0\)

\(\Rightarrow x\left(x-\frac{2006}{7}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{2006}{7}=0\end{cases}}\)

\(\Rightarrow x=\orbr{\begin{cases}0\\\frac{2006}{7}\end{cases}}\)

b) \(5.\left(x-2\right)+3x\left(2-x\right)=0\)

\(=5\left(x-2\right)-3x\left(x-2\right)=0\)

\(=\left(x-2\right)\left(5-3x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\5-3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\3x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{3}\end{cases}}\)

Ta có: \(\hept{\begin{cases}|x+\frac{2006}{2007}|\ge0;\forall x,y\\|\frac{2008}{2009}-y|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow|x+\frac{2006}{2007}|+|\frac{2008}{2009}-y|\ge0;\forall x,y\)

Do đó : \(\Rightarrow|x+\frac{2006}{2007}|+|\frac{2008}{2009}-y|=0\)

\(\Leftrightarrow\hept{\begin{cases}|x+\frac{2006}{2007}|=0\\|\frac{2008}{2009}-y|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2006}{2007}\\y=\frac{2008}{2009}\end{cases}}\)

Vậy ...

a) \(A=x^{15}+3x^{14}+5\)

\(=x^{14}\left(x+3\right)+5\)

\(=x^{14}.0+5\)

= 5

b) x = -3 => x + 3 = 0

\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(=\left(x^{2006}.0+1\right)^{2007}\)

\(=1^{2007}=1\)

\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)

\(A=x^{14}.\left(x+3\right)+5\)

\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)

\(A=0+5\)

\(A=5\)        \(\text{Vậy }A=5\text{ với x+3=0}\)

\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)

\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)

\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=\left(x^{2006}.0+1\right)^{2007}\)

\(B=\left(0+1\right)^{2007}\)

\(B=1^{2007}\)

\(B=1\)           \(\text{Vậy }B=1\text{ với x=-3}\)

Ngu thì câm mồm đi

Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

(x+2006/2007)⁶=0

=>x+2006/2007=0

x=0-2006/2007

x=-2006/2007

\(\lim\limits_{x\rightarrow+\infty}\dfrac{m+\dfrac{2006}{x}}{1+\sqrt{1+\dfrac{2007}{x^2}}}=\dfrac{m}{2}\)

\(A=0\Leftrightarrow\dfrac{m}{2}=0\Rightarrow m=0\)

minh moi hok lop 6