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a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)
\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)
c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)
d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)
\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a: Ta có: \(4x^2+12x+1\)
\(=4x^2+12x+9-8\)
\(=\left(2x+3\right)^2-8\ge-8\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
b: Ta có: \(4x^2-3x+10\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)
c: Ta có: \(2x^2+5x+10\)
\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)
\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)
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\(A=\left(2x\right)^2-2\times2x\times3+9+1\)
\(A=\left(2x-3\right)^2+1\)
Nhận xét:
\(\left(2x-3\right)^2\ge0\)
\(=>\left(2x-3\right)^2+1\ge1\)
\(=>A\ge1\)
Vậy A đạt GTNN tại A=1 <=> x=3/2
A = 4x2 -12x + 10
= (2x)2 - 2.2x.3 + 32 + 1
= (2x -3)2 +1 >= 1 với mọi x
Min A = 1 khi (2x -3)2 =0
<=> 2x - 3 = 0
<=> 2x = 3
<=> x = 3/2
Vậy Min A=1 khi x = 3/2
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`a)x^2-2x+2+4y^2+4y`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
`b)4x^2+y^2+12x+4y+13`
`=4x^2+12x+9+y^2+4y+4`
`=(2x+3)^2+(y+2)^2`
`c)x^2+17+4y^2+8x+4y`
`=x^2+8x+16+4y^2+4y+1`
`=(x+4)^2+(2y+1)^2`
`d)4x^2-12xy+y^2-4y+13`
`=4x^2-12x+9+y^2-4y+4`
`=(2x-3)^2+(y-2)^2`
a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)
b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)
c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)
d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
Mấy cái kia tương tự
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\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
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a: ta có: \(A=x^2-3x+10\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}>0\forall x\)
b: Ta có: \(B=x^2-5x+2021\)
\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{8015}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{8015}{4}>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn coi lại đề, GTLN và GTNN của biểu thức \(\dfrac{4x^2-8x+5}{x^2+1}\) rất xấu, và phải dùng kiến thức lớp 9 để tìm
vâng bn có thể lm kiến thức lớp 9 về delta để giải hộ m dc ko akk
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\(A=x^2+x\) . Có: \(x^2\ge x\Rightarrow x^2+x\ge0\)
Dấu '=' xảy ra khi: \(x^2+x=0\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy: \(Min_A=0\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(B=4x-12x+10\)
\(B=-8x+10\)
\(B=10-8x\)
Xét: \(x< 0\Rightarrow10-8x\ge10\)
Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)
Xét: \(x>0\Rightarrow10-8x\le10\)
Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)
Vậy: Khi x<0. \(Min_B=10\) tại \(x=0\)
Khi: x>0. \(Max_B=10\)tại \(x=0\)
K chắc
Ta có : \(4x^2-12x+10=\left(2x\right)^2-2.2x.3+9+1\)
\(=\left(2x-3\right)^2+1\ge1\) vì \(\left(2x-3\right)^2\ge0\)
Amin = 1 \(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy GTNN của A là 1 \(\Leftrightarrow x=\dfrac{3}{2}\)
Ta có: 4x2-12x+10= (2x-3)2+1
vì (2x-3)2\(\ge\)0
nên (2x-3)2+1>1
dấu bằng xảy ra khi: 2x-3=0
\(\Leftrightarrow\)x=\(\dfrac{3}{2}\)
Vậy GTNN của A là 1 khi x=\(\dfrac{3}{2}\)