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\(2x^2+6x-5=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{19}{2}\ge-\dfrac{19}{2}\)
Dấu "=" xảy ra khi \(x=-\dfrac{3}{2}\)
\(x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm giá trị nhỏ nhất của biểu thức:
a) Ta có:
\(M=2x^2+4x+7\)
\(M=2\cdot\left(x^2+2x+\dfrac{7}{2}\right)\)
\(M=2\cdot\left(x^2+2x+1+\dfrac{5}{2}\right)\)
\(M=2\cdot\left[\left(x+1\right)^2+2,5\right]\)
\(M=2\left(x+1\right)^2+5\)
Mà: \(2\left(x+1\right)^2\ge0\forall x\) nên:
\(M=2\left(x+1\right)^2+5\ge5\forall x\)
Dấu "=" xảy ra:
\(2\left(x+1\right)^2+5=5\Leftrightarrow2\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy: \(M_{min}=5\) khi \(x=-1\)
b) Ta có:
\(N=x^2-x+1\)
\(N=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\) nên \(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=" xảy ra:
\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(N_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
Tìm giá trị lớn nhất của biểu thức
a) Ta có:
\(E=-4x^2+x-1\)
\(E=-\left(4x^2-x+1\right)\)
\(E=-\left[\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right]\)
\(E=-\left[\left(2x-\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\)
Mà: \(\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\ge\dfrac{15}{16}\forall x\) nên
\(\Rightarrow E=-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\le-\dfrac{15}{16}\forall x\)
Dấu "=" xảy ra:
\(-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]=-\dfrac{15}{16}\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2-\dfrac{15}{16}=-\dfrac{15}{16}\)
\(\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2=0\Leftrightarrow2x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)
Vậy: \(E_{max}=-\dfrac{15}{16}\) khi \(x=\dfrac{1}{16}\)
b) Ta có:
\(F=5x-3x^2+6\)
\(F=-3x^2+5x-6\)
\(F=-\left(3x^2-5x-6\right)\)
\(F=-3\left(x^2-\dfrac{5}{3}x-2\right)\)
\(F=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{97}{36}\right]\)
\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\)
Mà: \(-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\) nên:
\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\le\dfrac{97}{36}\forall x\)
Dấu "=" xảy ra:
\(-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}=\dfrac{97}{36}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)
Vậy: \(F_{max}=\dfrac{97}{36}\) khi \(x=\dfrac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A=2x2-3x+1=\(2\left(x^2-2.\dfrac{3}{4}+\dfrac{9}{16}\right)-\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)
Vì \(2\left(x-\dfrac{3}{4}\right)^2\ge0\)
\(\Rightarrow A\ge-\dfrac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)
Vậy,Min \(A=\dfrac{-1}{8}\Leftrightarrow x=\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ta có: \(P=x^2+10x+27\)
\(=x^2+10x+25+2\)
\(=\left(x+5\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-5
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2 x 2 + 10 - 1 = 2 x 2 + 5 x - 1 / 2 B = 2 x 2 + 2 . 5 / 2 x + 5 / 2 2 - 5 / 2 2 - 1 / 2 = 2 x + 5 / 2 2 - 25 / 4 - 2 / 4 = 2 x + 5 / 2 2 - 27 / 2 = 2 x + 5 / 2 2 - 27 / 2 V ì x + 5 / 2 2 ≥ 0 n ê n 2 x + 5 / 2 2 ≥ 0 ⇒ 2 x + 5 / 2 2 - 27 / 2 ≥ - 27 / 2
Suy ra: B ≥ - 27/2 .
B= -27/2 khi và chỉ khi x + 5/2 = 0 suy ra x = -5/2
Vậy B = -27/2 là giá trị nhỏ nhất tại x = - 5/2
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\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+4\\ A=\left(x-y\right)^2+\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\Leftrightarrow x=y=1\)
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\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
\(A=2.\left(x^2-2.\frac{1}{4}.x+\frac{1}{16}\right)-\frac{9}{8}=2.\left(x-\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)