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

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

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

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

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

b, P = 2016 tại x = 2 và y = 0

cach lam nhu the nao 

a)x=9

d)x=-12;-6;-4;2

khẩn cấp thì ghi đáp án cho nhanh

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