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B=\(4x^2-4x+1+x^2+4x+4=5x^2+5\)
\(=5\left(x^2+1\right)\)
vì\(x^2+1\ge1\forall x\)
\(\Leftrightarrow B\ge5\forall x\)
dấu'=' xảy ra \(\Leftrightarrow x^2+1=0\Leftrightarrow x=0\)
vậy B đạt GTNN =5 khi x=0
Bài 2:
a) Ta có: \(A=x^2-3x+5\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
Ta có: \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{3}{2}=0\)
hay \(x=\dfrac{3}{2}\)
Vậy: Giá trị nhỏ nhất của biểu thức \(A=x^2-3x+5\) là \(\dfrac{11}{4}\) khi \(x=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x^2+x+1\right|+\left|x^2+3x+7\right|\ge\left|2x^2+4x+8\right|\)
Để đạt Min thì:
A=\(2x^2+4x+8\)
\(A=\left(\sqrt{2x}+\sqrt{2}\right)^2+6\)
\(\Rightarrow Min_A=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{8x^2-24x+32}{8\left(x-1\right)^2}=\frac{x^2-10x+25+7\left(x-1\right)^2}{8\left(x-1\right)^2}=\frac{\left(x-5\right)^2}{8\left(x-1\right)^2}+\frac{7}{8}\ge\frac{7}{8}\forall x\)
Dấu "=" xảy ra khi \(x-5=0\Rightarrow x=5\)
Vậy GTNN của A là \(\frac{7}{8}\) khi x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=\left(x-1\right)^4+\left(x-3\right)^4+6\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left(x-1\right)^4+2\left(x-1\right)^2\left(x-3\right)^2+\left(x-3\right)^4+4\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left[\left(x-1\right)^2+\left(x-3\right)^2\right]^2+4\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left[2x^2-8x+10\right]^2+4\left(x^2-4x+3\right)^2\)
\(A=\left[2\left(x-2\right)^2+2\right]+4\left[\left(x-2\right)^2-1\right]^2\)
\(A=4\left(x-2\right)^4+8\left(x-2\right)^2+4+4\left(x-2\right)^4-8\left(x-2\right)^2+4\)
\(A=8\left(x-2\right)^4+8\ge8\)
Vậy GTNN của biểu thức A là 8 \(\Leftrightarrow x=2\)
Đặt x-2=y
=> \(A=\left(y+1\right)^4+\left(y-1\right)^4+6\left(y+1\right)^2\left(y-1\right)^2\)
Khai triển A ta được
\(A=2y^4+12y^2+2+6\left(y^4-2y^2+1\right)\)
\(=8y^4+8=8\left(y^4+1\right)\ge8\)
Dấu "=" xảy ra khi y=0 lúc đó x=0+2=2
Vậy Amin=8 khi x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(x+1\right)\left(x+5\right)\left(x-2\right)\left(x+8\right)\)
\(=\left(x^2+6x+5\right)\left(x^2+6x-16\right)\)
\(=\left(x^2+6x-16\right)^2+21\left(x^2+6x-16\right)\)
\(=\left(x^2+6x-16+\frac{21}{2}\right)^2-\frac{441}{4}\ge-\frac{441}{4}\)
\(P_{min}=-\frac{441}{4}\) khi \(x^2+6x-16+\frac{21}{2}=0\)
\(Q=\left(x^2+\frac{y^2}{4}+\frac{9}{4}+xy-3x-\frac{3}{2}y\right)+\frac{3}{4}\left(y^2-2y+1\right)+2017\)
\(Q=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+\frac{3}{4}\left(y-1\right)^2+2017\ge2017\)
\(Q_{min}=2017\) khi \(x=y=1\)
Tìm GTLN, GTNN của biểu thức sau
\(1,A=\left(x-1\right)^2-10\)
\(2,B=-|x-1|-2\left(2y-1\right)^2+100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(A=\left(x-1\right)^2-10\ge-10\)
Dấu '=' xảy ra khi x=1
2: \(B=-\left|x-1\right|-2\cdot\left(2y-1\right)^2+100\le100\)
Dấu '=' xảy ra khi x=1 và y=1/2
`(x-1)^2 >=0 => (x-1)^2 - 10 >= -10`
Dấu bằng xảy ra khi `x = 1`.
Vì `-|x-1| <=0, -2(2y-1)^2 <= 0`
`=> -|x-1| - 2(2y-1)^2 + 100 <= 100`
Dấu bằng xảy ra `<=> x = 1, y = 1/2`.