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\(E=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\Rightarrow E_{min}=-\frac{5}{4}\) khi \(x=-\frac{3}{2}\)
\(F=\left(x^2+5x+4\right)\left(x^2+5x+6\right)=\left(x^2+5x+4\right)+2\left(x^2+5x+4\right)+1-1\)
\(F=\left(x^2+5x+5\right)^2-1\ge-1\)
\(\Rightarrow E_{min}=-1\) khi \(x^2+5x+5=0\Rightarrow x=\frac{-5\pm\sqrt{5}}{2}\)
\(M=\frac{2}{-4-\left(3x-1\right)^2}\ge\frac{2}{-4}=-\frac{1}{2}\Rightarrow M_{min}=-\frac{1}{2}\) khi \(x=\frac{1}{3}\)
\(P=\frac{x^2+2x+3}{x^2+2}\Rightarrow Px^2+2P=x^2+2x+3\)
\(\Rightarrow\left(P-1\right)x^2-2x+2P-3=0\)
\(\Delta'=1-\left(P-1\right)\left(2P-3\right)\ge0\)
\(\Leftrightarrow-2P^2+5P-2\ge0\Rightarrow\frac{1}{2}\le P\le2\)
\(\Rightarrow P_{max}=2\) khi \(x=1\)
\(P_{min}=\frac{1}{2}\) khi \(x=-2\)
a)
\(\dfrac{1}{x^2-2x+3}=\dfrac{1}{x^2-2x+1+2}=\dfrac{1}{\left(x-1\right)^2+2}\ge\dfrac{1}{2}\)
=> Min = \(\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(x=1\)
b) Viết lại đề
Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
\(a.M=x^2+6x-5\\ x^2+2\cdot3\cdot x+9-14\\ =\left(x+3\right)^2-14\ge-14\\ \Rightarrow GTNN\left(M\right)=-14\Leftrightarrow x+3=0\Rightarrow x=-3\)
câu b thấy thiếu thiếu x . y
\(x^2+2x+3\\ =x^2+2x+1+2\\ =\left(x^2+2x+1\right)+2\\ =\left(x+1\right)^2+2\\ Do\text{ }\left(x+1\right)^2\ge0\forall x\\ \Rightarrow\left(x+1\right)^2+2\ge2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\\ \text{Vậy }GTNN\text{ của biểu thức là }2\text{ }khi\text{ }x=-1\)
\(A=x^2+2x+3\)
\(A=x^2+2x+1+2\)
\(A=\left(x^2+2x+1\right)+2\)
\(A=\left(x+1\right)^2+2\)
Vậy GTNN của A=2 khi x=-1
d)
\(A=x^2-4x+2019=x^2-4x+4+2015=\left(x-2\right)^2+2015\ge2015\)
MinA = 2015 khi x = 2
a ) ĐKXĐ : \(x\ne-1\)
\(P=\dfrac{2x^2+3x+1}{x^3+x^2+2x+2}=\dfrac{2x\left(x+1\right)+x+1}{x^2\left(x+1\right)+2\left(x+1\right)}=\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x^2+2\right)\left(x+1\right)}=\dfrac{2x+1}{x^2+2}\)
b ) Tìm Min
\(P+\dfrac{1}{2}=\dfrac{2x+1}{x^2+2}+\dfrac{1}{2}=\dfrac{4x+2+x^2+2}{2\left(x^2+2\right)}=\dfrac{\left(x+2\right)^2}{2\left(x^2+2\right)}\)
\(\Rightarrow P=\dfrac{\left(x+2\right)^2}{x^2+2}-\dfrac{1}{2}\ge-\dfrac{1}{2}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x=-2\)
Tìm Max
\(P-1=\dfrac{2x+1}{x^2+2}-1=\dfrac{2x+1-x^2-2}{x^2+2}=\dfrac{-\left(x-1\right)^2}{x^2+2}\)
\(\Rightarrow P=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x=1\)
Vậy ...
P/s : Sử dụng Delta để làm nhé bạn :D
\(a)P=\dfrac{2x^2+3x+1}{x^3+x^2+2x+2}\\ P=\dfrac{2x^2+2x+x+1}{x^2\left(x+1\right)+2\left(x+1\right)}\\ P=\dfrac{2x\left(x+1\right)+x+1}{\left(x+1\right)\left(x^2+2\right)}\\ P=\dfrac{\left(x+1\right)\left(2x+1\right)}{\left(x+1\right)\left(x^2+2\right)}\\ P=\dfrac{2x+1}{x^2+2}\)
a) Áp dụng BĐT bunhiacopxki ta có:
A= \(a^2+b^2\) \(\geq\) \(\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
Vậy Min A= \(\dfrac{1}{2}\) khi a=b=\(\dfrac{1}{2}\)
b) Ta có: B= \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\)
\(\Leftrightarrow\) B= \(\left(\dfrac{a^2}{b}+b\right)+\left(\dfrac{b^2}{a}+a\right)-\left(a+b\right)\) \(\geq\) \(2\sqrt{\dfrac{a^2}{b}.b}+2\sqrt{\dfrac{b^2}{a}.a}-a-b\) = \(2a+2b-a-b\) \(=a+b=1\)
Từ đó suy ra: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\) \(\geq\) 1
Vậy Min B = 1 khi a=b=\(\dfrac{1}{2}\)
\(B=\left(x-1\right)^2-4\ge4\\ B_{min}=4\Leftrightarrow x=1\)
\(B=x^2-2x-3=\left(x^2-2x+1\right)-4\)
\(=\left(x-1\right)^2-4\ge-4\)
\(minB=-4\Leftrightarrow x=1\)