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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: ac giúp e mấy câu này vs, lát nx e nộp r Bài 16. Tìm GTNN của A = (x+1)² + (x-2)² +x(x-1) -1 Bài 18. Tìm GTNN của A = (2x-1) (x+1) -1 B

Toán Lớp 8: ac giúp e mấy câu này vs, lát nx e nộp r
Bài 16. Tìm GTNN của A = (x+1)² + (x-2)² +x(x-1) -1
Bài 18. Tìm GTNN của A = (2x-1) (x+1) -1
Bài 19. Tìm GTNN của A = (3x+ 1).(x-2) -20
Bài 20. Tìm GTNN của A = (2x+1)² – x(x-1) -1

Comments ( 1 )

  1. Giải đáp:
    $16)min_ A=\dfrac{13}{4} \Leftrightarrow x=\dfrac{1}{2}\\ 18) min_ A -\dfrac{17}{8} \Leftrightarrow x=\dfrac{1}{4}\\ 19) min_ A =  -\dfrac{289}{12} \Leftrightarrow x=\dfrac{5}{6}\\ 20)min_ A =\dfrac{25}{12} \Leftrightarrow x=-\dfrac{5}{6}$
    Lời giải và giải thích chi tiết:
    $16)\\ A=(x+1)^2 + (x-2)^2 +x(x-1) -1\\ =x^2+2x+1 + x^2-4x+4 +x^2-x -1\\ =3x^2-3x+4\\ =3\left(x^2-x+\dfrac{4}{3}\right)\\ =3\left(x^2-x+\dfrac{1}{4}+\dfrac{13}{12}\right)\\ =3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{13}{4}\\ =3\left(x-\dfrac{1}{2}\right)^2+\dfrac{13}{4} \ge \dfrac{13}{4} \ \forall \ x$
    Dấu “=” xảy ra $\Leftrightarrow x-\dfrac{1}{2} =0 \Leftrightarrow x=\dfrac{1}{2}$
    $18)\\ A = (2x-1) (x+1) -1\\ =2x^2+2x-x-1-1\\ =2x^2+x-2\\ =2\left(x^2+\dfrac{x}{2}-1\right)\\ =2\left(x^2+2.\dfrac{x}{4}+\dfrac{1}{16}-\dfrac{17}{16}\right)\\ =2\left(x^2+2.\dfrac{x}{4}+\dfrac{1}{16}\right)-\dfrac{17}{8}\\ =2\left(x-\dfrac{1}{4}\right)^2-\dfrac{17}{8} \ge -\dfrac{17}{8} \ \forall \ x$
    Dấu “=” xảy ra $\Leftrightarrow x-\dfrac{1}{4} =0 \Leftrightarrow x=\dfrac{1}{4}$
    $19)\\ A = (3x+ 1)(x-2) -20\\ =3x^2-6x+x-2-20\\ =3x^2-5x-22\\ =3\left(x-\dfrac{5}{3}x-\dfrac{22}{3}\right)\\ =3\left(x-2.\dfrac{5}{6}x+\dfrac{25}{36}-\dfrac{289}{36}\right)^2\\ =3\left(x-2.\dfrac{5}{6}x+\dfrac{25}{36}\right)^2-\dfrac{289}{12}\\ =3\left(x-\dfrac{5}{6}\right)^2-\dfrac{289}{12} \ge -\dfrac{289}{12} \ \forall \ x$
    Dấu “=” xảy ra $\Leftrightarrow x-\dfrac{5}{6}=0 \Leftrightarrow x=\dfrac{5}{6}$
    $20)\\ A = (2x+1)^2 – x(x-1) -1\\ =4x^2+4x+1-x^2+x-1\\ =3x^2+5x\\ =3\left(x+\dfrac{5}{3}x\right)\\ =3\left(x+2.\dfrac{5}{6}x+\dfrac{25}{36}-\dfrac{25}{36}\right)^2\\ =3\left(x+2.\dfrac{5}{6}x+\dfrac{25}{36}\right)^2-\dfrac{25}{12}\\ =3\left(x+\dfrac{5}{6}\right)^2-\dfrac{25}{12} \ge -\dfrac{25}{12} \ \forall \ x$
    Dấu “=” xảy ra $\Leftrightarrow x+\dfrac{5}{6}=0 \Leftrightarrow x=-\dfrac{5}{6}$

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222-9+11+12:2*14+14 = ? ( )