a) Ta có: \(2x+x^2=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b) Ta có: \(\left(2x+1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

đặt cái chung ra ngoài

a) \(5x+10y=5\left(x+2y\right)\)

b) \(3x^2y+9xy^2z=3xy\left(x+3yz\right)\)

g) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)

h) \(x^2+9x+8=\left(x+8\right)\left(x+1\right)\)

l) \(x^2-10x+9=\left(x-1\right)\left(x-9\right)\)

k) \(x^2+x-12=\left(x+4\right)\left(x-3\right)\)

l) \(3x^2+8x+4=\left(3x+2\right)\left(x+2\right)\)

bn tách ra đi

a)Tỉ lệ KG đồng hợp : AA = aa  \(\dfrac{1-\left(\dfrac{1}{2}\right)^3}{2}=\dfrac{7}{16}\)

b) tỉ lệ KG dị hợp :  \(\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\)

c) _{bn ghi F mấy ko rõ nên mik xin lm F4} :

Cho F3 tự thụ phấn :

\(\dfrac{7}{16}\left(AAxAA\right)->F4:\dfrac{7}{16}AA\)

\(\dfrac{1}{8}\left(AaxAa\right)->F4:\dfrac{1}{32}AA:\dfrac{2}{32}Aa:\dfrac{1}{32}aa\)

\(\dfrac{7}{16}\left(aaxaa\right)->F4:\dfrac{7}{16}aa\)

Cộng các Kquả lại ta đc :

F4 : KG :    \(\dfrac{15}{32}AA:\dfrac{2}{32}Aa:\dfrac{15}{32}aa\)

      KH :  \(\dfrac{17}{32}trội:\dfrac{15}{32}lặn\)

(còn nếu đề mak ghi lak thế hệ F1 thik chỉ cần lm sđlai Aa x Aa như thường thôi nha :v )

Sao không áp dụng CT của câu a,b cho câu c luôn nếu là F4 . Dài dòng quá!

a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

\(Q=\dfrac{x\left(-x^2-2x-6\right)}{x}\)

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

\(Q=-\left(x^2+2x+6\right)\)

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)

d: \(\Leftrightarrow x+5=5\)

hay x=0