a) Theo đề bài: %X- %T= 20%N
Theo hệ quả NTBS ta có: %T+ %X= 50%N
=> \(\left\{{}\begin{matrix}\%X-\%T=20\%N\\\%X+\%T=50\%\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%X=\%G=35\%N\\\%T=\%A=15\%N\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\%G+\%T=50\%N\\\frac{\%G}{\%T}=\frac{3}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%G=\%X=15\%N\\\%A=\%T=35\%N\end{matrix}\right.\)
Theo bài ra ta có:
$A × G = 5,25\%; A + G = 50\%$
$→ A × (50\% - A) = 5,25\%$
Giải phương trình ta được:
$A = 35\% → G = X = 15\%$
Hoặc: $A = 15\% → G = X = 35\%$
c) \(\left\{{}\begin{matrix}\%A+\%X=50\%N\\\%A.\%X=5,25\%N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=0,5N-X\\\left(0,5N-X\right).X=0,0525N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}A=0,5N-X\\0,5X-X^2-0,0525N=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}A=0,5N-X\\\left[{}\begin{matrix}X=0,35N\\X=0,15N\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}G=X=0,35N=35\%N\\A=T=0,15N=15\%N\end{matrix}\right.\\\left\{{}\begin{matrix}G=X=0,15N=15\%N\\A=T=0,35N=35\%N\end{matrix}\right.\end{matrix}\right.\)
=> Có 2 TH:
TH1: %G=%X=35%N -> %A=%T=15%N
TH2: %G=%X=15%N -> %A=%T= 35%N
d) GEN 4:
\(\left\{{}\begin{matrix}A^2-X^2=0,15N\\A+X=0,5N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(A-X\right).\left(A+X\right)=0,15N\\A+X=0,5N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(A-X\right).0,5=0,15N\\A+X=0,5N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}A-X=0,3N\\A+X=0,5N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}A=T=0,4N=40\%N\\G=X=0,1N=10\%N\end{matrix}\right.\)
e) GEN 5:
\(\left\{{}\begin{matrix}X+T=0,5N\\X^2+T^2=0,15N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}X=0,5-T\\\left(0,5-T\right)^2+T^2-0,15=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}X=0,5-T\\\left(0,25-T+T^2\right)+T^2-0,15=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}X=0,5-T\\2T^2-T+0,1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}X=0,5-T\\\left[{}\begin{matrix}T=X\approx36,18\%N\\T=X\approx13,82\%N\end{matrix}\right.\end{matrix}\right.\)
Vậy có 2 TH xảy ra:
TH1: T=X\(\approx\) 36,18% N -> G=X\(\approx\) 13,82%N
TH2: T=X\(\approx\) 13,82%N -> G=X\(\approx\) 36,18%N
f) GEN 6
\(\left\{{}\begin{matrix}T^3+G^3=0,04625N\\T+G=0,5N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\T^3+G^3=0,04625N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\\left(T+G\right).\left(T^2-T.G+G^2\right)=0,04625N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\0,5.\left(T^2-T.G+G^2\right)=0,04625N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\T^2-T.G+G^2=0,0925N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\\left(T+G\right)^2-3.T.G=0,0925N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\\left(0,5^2-0,0925\right)N=3.T.G\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\T.G=\frac{0,1575}{3}N=0,0525N\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\\left(0,5-G\right).G-0,0525=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}T=0,5-G\\-G^2+0,5.G-0,0525=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}A=T=0,5N-G\\\left[{}\begin{matrix}X=G=0,35N\\X=G=0,15N\end{matrix}\right.\end{matrix}\right.\)
Vậy GEN 6 có 2TH:
TH1: G=X=35%N -> A=T=15%N
TH2: G=X=15%N-> A=T=35%N