DK:  \(x\ne-3\)

\(y=\frac{x^2-9}{x+3}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3\)

Nhu vay hso da cho la ham so bac nhat

à tiện thể

vì quá chán nên chat vs tôi luôn nhé

\(\text{rất vui vì bạn chưa chết}\)

viết lung tung 

Tổng số nu của gen

Nu = 150 . 20 = 3000 (nu)

Theo đề : A - G = 10%

Lại có      A + G = 50 % 

=> A = T = 30 % ; G= X = 20%

\(A=T=3000.30\%=900\left(nu\right)\)

\(G=X=3000.20\%=600\left(nu\right)\)

b)  TH1 : Thay 2 cặp A-T bằng 2 cặp G-X

=> Dạng : Thay cặp nu này = cặp nu khác

TH2 : Thêm 1 cặp A-T

=> Dạng : Thêm 1 cặp nu

\(x^5-5x^4+4x^3+4x^2-5x+1=0\)

\(\left(x^5-x^4\right)-\left(4x^4-4x^3\right)+\left(4x^2-4x\right)-\left(x-1\right)=0\)

\(x^4\left(x-1\right)-4x^3\left(x-1\right)+4x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^4-4x^3+4x-1\right)=0\)

\(\left(x-1\right)\left[\left(x^4-1\right)-\left(4x^3-4x\right)\right]=0\)

\(\left(x-1\right)\left[\left(x-1\right)\left(x^3+x^2+x+1\right)-4x\left(x^2-1\right)\right]=0\)

\(\left(x-1\right)\left[\left(x-1\right)\left(x^3+x^2+x+1\right)-4x\left(x-1\right)\left(x+1\right)\right]=0\)

\(\left(x-1\right)^2\left(x^3+x^2+x+1-4x^2-4x\right)=0\)

\(\left(x-1\right)^2\left(x^3-3x^2-3x+1\right)=0\)

\(\left(x-1\right)^2\left[\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left(x^2-x+1-3x\right)=0\)

\(\left(x-1\right)^2\left(x+1\right)\left[\left(x^2-2.x.2+2^2\right)-3\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left[\left(x-2\right)^2-\left(\sqrt{3}\right)^2\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)

Đến đây b tự làm tiếp nhé~

\(x^2-y^2+x-y=5\)\(\Leftrightarrow\left(x^2-y^2\right)+\left(x-y\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+\left(x-y\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x-y+1\right)=5\)

\(x^3-x^2y-xy^2+y^3=6\)

\(\Leftrightarrow\left(x^3+y^3\right)-\left(x^2y+xy^2\right)=6\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)=6\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2-xy\right)=6\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-2xy+y^2\right)=6\)

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