1/ Tìm GTNN của:
a/ \(f\left(x\right)=5x^2-2x+1\)
b/ \(P\left(x\right)=3x^2+x+7\)
c/ \(Q\left(x\right)=5x^2-3x-3\)
2/ Tìm GTLN của:
a/ \(f\left(x\right)=-3x^2+x-2\)
b/ \(P\left(x\right)=-x^2-7x+1\)
c/ \(Q\left(x\right)=-2x^2+x-8\)
3/ Cho \(\hept{\begin{cases}x+y+z=0\\xy+yz+xz=0\end{cases}}\). CMR: x = y = z
4/ Chứng minh rằng biểu thức sau viết được dưới dạng tổng các bình phương của hai biểu thức:\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)
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 ...
3/
Ta có x + y + z = 0
<=> (x + y + z)2 = 0
<=> x2 + y2 + z2 + 2xy + 2xz + 2yz = 0
<=> x2 + y2 + z2 + 2(xy + xz + yz) = 0
Thay xy + yz + xz = 0 vào BT trên
=> x2 + y2 + z2 = 0
Mà \(x^2\ge0;y^2\ge0;z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)
=> x = y = z (đpcm)
3.
Theo đề ra ta có :
\(\hept{\begin{cases}x+y+z=0\\xy+yz+xz=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\2\left(xy+yz+xz\right)=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2+y^2+z^2+2xy+2yz+2zx=0\left(1\right)\\2xy+2yz+2zx=0\left(2\right)\end{cases}}\)
Lấy (1) trừ (2) ta được :
\(x^2+y^2+z^2=0\)
Lại có : \(\hept{\begin{cases}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{cases}}\)
\(\Rightarrow x^2+y^2+z^2\ge0\)
Dấu "=" xảy ra khi :
x = y = z = 0
Vậy ...
