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Ta có: F = (x - 2)(3x + 1) + 5 = 3x2 + x - 6x - 2 + 5 = 3x2 - 5x + 3 = 3(x2 - 5/3x + 25/36) + 11/12 = 3(x - 5/6)2 + 11/12
Ta luôn có: (x - 5/6)2 \(\ge\)0 \(\forall\)x --> 3(x - 5/6)2 \(\ge\) 0 \(\forall\)x
=> 3(x - 5/6)2 + 11/12 \(\ge\)11/12\(\forall\)x
Hay F \(\ge\) 11/12 \(\forall\)x
Dấu "=" xảy ra khi : x - 5/6 = 0 <=> x = 5/6
Vậy Fmin = 11/12 tại x = 5/6
F = (x-2)(3x+1) + 5
F = 3x2 - 6x + x - 2 + 5
F = 3x2 - 5x + 3
F = 3(x2 - 5x) + 3
F = 3(x2 - 2.x.5/2 + 25/4 - 25/4) + 3
F = 3.(x-5/2)2 - 75/4 + 3
F = 3(x-5/2)2 -63/12 >= -63/12
Dấu ''='' xảy ra <=> x = 5/2
Vậy ...
\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
a) \(A=5-8x-x^2=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+8x+16-21\right)\)
\(=-\left[\left(x+4\right)^2-21\right]\)
\(=-\left(x+4\right)^2+21\le21\)
Vậy \(A_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
\(B=5x-3x^2=-3\left(x^2-\frac{5}{3}x\right)\)
\(=-3\left(x^2-\frac{5}{3}x+\frac{35}{36}-\frac{25}{36}\right)\)
\(=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{25}{36}\right]\)
\(=-3\left[\left(x-\frac{5}{6}\right)^2\right]+\frac{25}{12}\le\frac{25}{12}\)
Vậy \(B_{min}=\frac{25}{12}\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)