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Ta có: \(2x^2+2x+5=2\left(x^2+x+\dfrac{5}{2}\right)=2\left[x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{9}{4}\right]=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
=> \(M=\dfrac{1}{2x^2+2x+5}\le\dfrac{1}{\dfrac{9}{2}}=\dfrac{2}{9}\forall x\)
Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy MaxM=\(\dfrac{2}{9}\) khi x=\(-\dfrac{1}{2}\)
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A = y^2 - 4y + 9 = y^2 - 4y + 4 + 5
= ( y - 2 )^2 + 5 >= 5
Dấu ''='' xảy ra khi y = 2
Vậy GTNN A là 5 khi y = 2
B = x^2 - x + 1 = x^2 - x + 1/4 + 3/4 = ( x - 1/2 )^2 + 3/4 >= 3/4
Dấu ''='' xảy ra khi x = 1/2
Vậy GTNN B là 3/4 khi x = 1/2
C = 2x^2 - 6x = 2 ( x^2 - 3x + 9 / 4 - 9/4 )
= 2 ( x - 3/2 )^2 - 9/2 >= -9/2
Dấu ''='' xảy ra khi x = 3/2
Vậy GTNN C là -9/2 khi x = 3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
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a: \(M=2x^2-4x+3\)
\(=2x^2-4x+2+1\)
\(=2\left(x^2-2x+1\right)+1\)
\(=2\left(x-1\right)^2+1>=1\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
b: \(N=x^2-4x+5+y^2+2y^2\)
\(=x^2-4x+4+3y^2+1\)
\(=\left(x-2\right)^2+3y^2+1>=1\forall x,y\)
Dấu '=' xảy ra khi x-2=0 và y=0
=>x=2 và y=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2 : Bạn cần thêm điều kiện a,b là các số không âm
Áp dụng bất đẳng thức Cosi, ta có : \(12=3a+5b\ge2.\sqrt{3a.5b}=2\sqrt{15ab}\Rightarrow ab\le\frac{6^2}{15}=\frac{12}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3a+5b=12\\3a=5b\\a,b\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=\frac{6}{5}\end{cases}}\)
Vậy Max B = \(\frac{12}{5}\Leftrightarrow\hept{\begin{cases}a=2\\b=\frac{6}{5}\end{cases}}\)
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a) \(2x^2-4x=0\)
\(2x\left(x-2\right)=0\)
TH1:2x=0⇒x=0
TH2:x-2=0⇒x=2
\(a,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow x-4=0\Leftrightarrow x=4\)
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\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
-2x2+4x+5
=-2x2+4x-2+7
=-2.(x2-2x+1)+7
=-2.(x-1)2+7 < hoặc =7
Dấu "=" xảy ra khi : x=1
Vậy Max của biểu thức trên là 7 tại x=1