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Ta có : \(P=2x^2-8x+1=2\left(x^2-4x\right)+1=2\left(x^2-4x+4-4\right)+1=2\left(x-2\right)^2-7\)
Vì \(2\left(x-2\right)^2\ge0\forall x\)
Nên : \(P=2\left(x-2\right)^2-7\ge-7\forall x\in R\)
Vậy \(P_{min}=-7\) khi x = 2
Lời giải:
\(\bullet\)Nếu \(x\geq \frac{1}{2}\Rightarrow K=x-\frac{1}{2}+\frac{3}{4}-x=\frac{1}{4}\)
\(\bullet\) Nếu \(x<\frac{1}{2}\Rightarrow K=\frac{1}{2}-x+\frac{3}{4}-x=\frac{5}{4}-2x\)
Vì \(x<\frac{1}{2}\Rightarrow \frac{5}{4}-2x>\frac{5}{4}-1=\frac{1}{4}\)
Do đó \(K_{\min}=\frac{1}{4}\)
Hàm hiển nhiên không có max. Xét hàm \(\frac{5}{4}-2x\), với giá trị của \(x<\frac{1}{2}\), càng nhỏ thì $K$ càng lớn đến dương vô cùng.
TH1:Nếu x-\(\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(\Rightarrow\)K=\(\left|\dfrac{1}{2}-\dfrac{1}{2}\right|+\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)
TH2:Nếu x-\(\dfrac{1}{2}>0\Rightarrow x>\dfrac{1}{2}\Rightarrow\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)
\(\Rightarrow K=x-\dfrac{1}{2}+\dfrac{3}{4}-x=\dfrac{1}{4}\)
TH3:Nếu \(x-\dfrac{1}{2}< 0\Rightarrow x< \dfrac{1}{2}\Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{2}-x\)
\(\Rightarrow K=\dfrac{1}{2}-x+\dfrac{3}{4}-x\)
\(\Rightarrow K=\dfrac{5}{4}-2x< \dfrac{1}{4}\)
Vậy Max K=\(\dfrac{1}{4}\Leftrightarrow x\ge\dfrac{1}{2}\)
|x-1/2| =x-1/2 khi x >= 1/2
=> Min K =1/4 khi x>=1/2
không có max
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
a/ Đặt: \(x+\frac{1}{x}=a\)
Ta có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)
\(x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)^2-2\)
\(=\left(a^3-3a\right)^2-2\)
\(\Rightarrow M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^6-\left(a^3-3a\right)^2+2-2}{a^3+a^3-3a}\)
\(=\frac{\left(a^3+a^3-3a\right)\left(a^3-a^3+3a\right)}{\left(a^3+a^3-3a\right)}=3a\)
\(=3.\left(x+\frac{1}{x}\right)=\frac{3x^2+3}{x}\)
b/ \(\frac{3x^2+3}{x}=3x+\frac{3}{x}\ge2.3=6\)
Đấu = xảy ra khi \(x=\frac{1}{x}\Leftrightarrow x=1\)
Có : (x+2017)^2 = x^2+4034x+2017^2 = (x^2-4034x+2017^2)+8068x = (x-2017)^2+8068x >= 8068x
=> D <= x/8068x = 1/8068
Dấu "=" xảy ra <=> x-2017=0 <=> x = 2017
Vậy Max của D = 1/8068 <=> x = 2017
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