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1. Biến đổi biểu thức Acap A𝐴:
   • Xét A=x+3x2+7cap A equals the fraction with numerator x plus 3 and denominator x squared plus 7 end-fraction𝐴=𝑥+3𝑥2+7.
   • Nhân chéo để đưa về phương trình bậc hai theo xx𝑥: Ax2+7A=x+3cap A x squared plus 7 cap A equals x plus 3𝐴𝑥2+7𝐴=𝑥+3.
   • Chuyển vế: Ax2−x+(7A−3)=0cap A x squared minus x plus open paren 7 cap A minus 3 close paren equals 0𝐴𝑥2−𝑥+(7𝐴−3)=0.
2. • Để phương trình có nghiệm xx𝑥, biệt thức Δcap deltaΔphải lớn hơn hoặc bằng 000.
   • Δ=(-1)2−4A(7A−3)≥0cap delta equals open paren negative 1 close paren squared minus 4 cap A open paren 7 cap A minus 3 close paren is greater than or equal to 0Δ=(−1)2−4𝐴(7𝐴−3)≥0.
   • 1−28A2+12A≥01 minus 28 cap A squared plus 12 cap A is greater than or equal to 01−28𝐴2+12𝐴≥0.
   • 28A2−12A−1≤028 cap A squared minus 12 cap A minus 1 is less than or equal to 028𝐴2−12𝐴−1≤0.
3. . Giải bất phương trình bậc hai theo Acap A𝐴:
   • Tìm nghiệm của phương trình 28A2−12A−1=028 cap A squared minus 12 cap A minus 1 equals 028𝐴2−12𝐴−1=0.
   • ΔA=(-12)2−4(28)(-1)=144+112=256cap delta sub cap A equals open paren negative 12 close paren squared minus 4 open paren 28 close paren open paren negative 1 close paren equals 144 plus 112 equals 256Δ𝐴=(−12)2−4(28)(−1)=144+112=256.
   • ΔA=256=16the square root of cap delta sub cap A end-root equals the square root of 256 end-root equals 16Δ𝐴√=256√=16.
   • A1=12−162×28=-456=−114cap A sub 1 equals the fraction with numerator 12 minus 16 and denominator 2 cross 28 end-fraction equals negative 4 over 56 end-fraction equals negative 1 over 14 end-fraction𝐴1=12−162×28=−456=−114.
   • A2=12+162×28=2856=12cap A sub 2 equals the fraction with numerator 12 plus 16 and denominator 2 cross 28 end-fraction equals 28 over 56 end-fraction equals 1 over 2 end-fraction𝐴2=12+162×28=2856=12.
   • Vì 28A2−12A−1≤028 cap A squared minus 12 cap A minus 1 is less than or equal to 028𝐴2−12𝐴−1≤0, nên −114≤A≤12negative 1 over 14 end-fraction is less than or equal to cap A is less than or equal to 1 over 2 end-fraction−114≤𝐴≤12.
4. . Xác định giá trị lớn nhất của Acap A𝐴và tìm xx𝑥:
   • Giá trị lớn nhất của Acap A𝐴là 121 over 2 end-fraction12.
   • Thay A=12cap A equals 1 over 2 end-fraction𝐴=12vào phương trình Ax2−x+(7A−3)=0cap A x squared minus x plus open paren 7 cap A minus 3 close paren equals 0𝐴𝑥2−𝑥+(7𝐴−3)=0.
   • 12x2−x+(7×12−3)=01 over 2 end-fraction x squared minus x plus open paren 7 cross 1 over 2 end-fraction minus 3 close paren equals 012𝑥2−𝑥+(7×12−3)=0.
   • 12x2−x+(72−62)=01 over 2 end-fraction x squared minus x plus open paren 7 over 2 end-fraction minus 6 over 2 end-fraction close paren equals 012𝑥2−𝑥+(72−62)=0.
   • 12x2−x+12=01 over 2 end-fraction x squared minus x plus 1 over 2 end-fraction equals 012𝑥2−𝑥+12=0.
   • Nhân với 222: x2−2x+1=0x squared minus 2 x plus 1 equals 0𝑥2−2𝑥+1=0.
   • (x−1)2=0open paren x minus 1 close paren squared equals 0(𝑥−1)2=0.
   • x=1x equals 1𝑥=1.

Đáp án là

121 over 2 end-fraction12

giá trị lớn nhất của \(A\) là \(\frac{1}{2}\) và nó xảy ra khi \(x = 1\).

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