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\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(\Leftrightarrow A=\left(x^2-x+6x-6\right)\left(x^2+2x+3x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy GTNN của A là : \(-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
4.
= x\(^2\)-2.\(\dfrac{5}{2}\)x+\(\dfrac{25}{4}\)-\(\dfrac{13}{4}\)
= (x-\(\dfrac{5}{2}\))\(^2\)-\(\dfrac{13}{4}\)lớn hơn hoặc bằng -\(\dfrac{13}{4}\) với mọi x
=> min= -\(\dfrac{13}{4}\) <=> x = 5/2
5.
= 2( x\(^2\)-\(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\))
=2( x\(^2\)-2.\(\dfrac{5}{4}\)+\(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))
=2( x-\(\dfrac{5}{4}\))\(^2\)-\(\dfrac{27}{2}\) lớn hơn hoặc bằng -27/2 với mọi x
vậy min = -\(\dfrac{27}{2}\) <=> x= 5/4
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
\(E=x^2-\frac{1}{2}x+1\)
\(=\left(x^2-\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
vậy min E=3/4 <=> x=1/2
1: \(=x^2+x+5=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}>=\dfrac{19}{4}\)
Dấu '=' xảy ra khi x=-1/2
2: \(=-\left(x^2+4x-9\right)\)
\(=-\left(x^2+4x+4-13\right)\)
\(=-\left(x+2\right)^2+13\le13\)
Dấu '=' xảy ra khi x=-2
3: \(=x^2-4x+4+y^2+2y+1+2\)
\(=\left(x-2\right)^2+\left(y+1\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2 và y=-1
Ta có:
A = -x2 - 4x - 2 = -(x2 + 4x + 4) + 2 = -(x + 2)2 + 2
Ta luôn có: -(x + 2)2 \(\le\)0 \(\forall\)x
=> -(x + 2)2 + 2 \(\le\)2 \(\forall\)x
Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2
Vậy Max của A = 2 tại x = -2
(xem lại đề)
\(A=36-3x+\dfrac{1}{2}x^2=\dfrac{1}{2}\left(x^2-6x+72\right)\)
\(=\dfrac{1}{2}\left[\left(x^2-6x+9\right)+63\right]=\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\)
Có: \(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+63\ge63\)
\(\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\ge\dfrac{1}{2}\cdot63=\dfrac{63}{2}\)
Dấu ''='' xảy ra khi x = 3
Vậy \(MIN_A=\dfrac{63}{2}\Leftrightarrow x=3\)