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\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
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 \(B_{min}=-8\Leftrightarrow x=2\)
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)
đẳng thức xảy ra khi :
\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
a, 4C = 12|x|+8/4|x|-5 = 3 + 23/|x|-5 <= 3 + 23/0-5 = -8/5
=> C <= -2/5
Dấu "=" xảy ra <=> x=0
Vậy Min ...
b, Để C thuộc N => 3|x|+2 chia hết cho 4|x|-5
=> 4.(3|x|+2) chia hết cho 4|x|-5
<=> 12|x|+8 chia hết cho 4|x|-5
<=> 3.(|x|+5) + 23 chia hết cho 4|x|-5
=> 23 chia hết chi 4|x|-5 [ vì 3.(4|x|-5) chia hết cho 4|x|-5 ]
Đến đó bạn tìm ước của 23 rùi giải
đề dài v~
1.
a) \(f\left(x\right)=5x^2-2x+1\)
\(5f\left(x\right)=25x^2-10x+5\)
\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)
\(5f\left(x\right)=\left(5x-1\right)^2+4\)
Mà \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow5f\left(x\right)\ge4\)
\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)
Dấu " = " xảy ra khi :
\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)
Vậy ....
b) \(P\left(x\right)=3x^2+x+7\)
\(3P\left(x\right)=9x^2+3x+21\)
\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)
\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)
Mà \(\left(3x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)
\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)
Dấu "=" xảy ra khi :
\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)
Vậy ...
c) \(Q\left(x\right)=5x^2-3x-3\)
\(5Q\left(x\right)=25x^2-15x-15\)
\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)
\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)
Mà \(\left(5x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)
\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)
Dấu "=" xảy ra khi :
\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)
Vậy ...
2.
a) \(f\left(x\right)=-3x^2+x-2\)
\(-3f\left(x\right)=9x^2-3x+6\)
\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)
\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)
Mà \(\left(3x-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)
\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)
Dấu "=" xảy ra khi :
\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
b) \(P\left(x\right)=-x^2-7x+1\)
\(-P\left(x\right)=x^2+7x-1\)
\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)
\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x+\frac{7}{2}\right)^2\ge0\)
\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)
\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)
Vậy ...
c) \(Q\left(x\right)=-2x^2+x-8\)
\(-2Q\left(x\right)=4x^2-2x+16\)
\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)
\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)
Mà : \(\left(2x-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)
\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)
Dấu "=" xảy ra khi :
\(2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
Vậy ...
Bài làm:
+ \(C=10\left(x^2-2\right)+5=10x^2-20+5=10x^2-15\ge-15\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(10x^2=0\Rightarrow x=0\)
Vậy \(Min\left(C\right)=-15\Leftrightarrow x=0\)
+ \(D=\left(7-x\right)\left(2x+1\right)=-2x^2+13x+7=-2\left(x^2-\frac{13}{2}x+\frac{169}{16}\right)-\frac{225}{8}\)
\(=-2\left(x-\frac{13}{4}\right)^2-\frac{225}{8}\le-\frac{225}{8}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(-2\left(x-\frac{13}{4}\right)^2=0\Rightarrow x=\frac{13}{4}\)
Vậy \(Max\left(D\right)=-\frac{225}{8}\Leftrightarrow x=\frac{13}{4}\)
+ \(H=x^2+y^2+2x-4y+10=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+5\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+5\ge5\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy \(Min\left(H\right)=5\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
+ \(E=-x^2-4x+6y-y^2-2021=-\left(x^2+4x+4\right)-\left(y^2-6y+9\right)-2008\)
\(=-\left(x+2\right)^2-\left(y-3\right)^2-2008\le-2008\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\left(x+2\right)^2=0\\-\left(y-3\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Vậy \(Max\left(E\right)=-2008\Leftrightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Học tốt!!!!
Ta có : \(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)\)
\(=\left(x^2+x\right)\left(x^2-3x+4x-12\right)\)
\(=\left(x^2+x\right)\left(x^2+x-12\right)\left(1\right)\)
Đặt \(x^2+x=t\)
\(\Rightarrow\left(1\right)\Leftrightarrow t\left(t-12\right)=t^2-12t=t^2-12t+36-36=\left(t-6\right)^2-36\)
Vì : \(\left(t-6\right)^2\ge0\)
\(\Rightarrow\left(t-6\right)^2-36\ge-36\)
Dấu " = " xảy ra khi \(t-6=0\)
\(t=0+6\)
\(t=6\)
\(\Rightarrow x^2+x+6\) \(x=2\) hoăc \(x=-3\)
Vậy \(MIN_B=-36\) khi \(x=2;x=-3\)
nhân lên r` đặt ẩn