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Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có : \(1=\frac{\left(\sqrt{a}\right)^2}{x}+\frac{\left(\sqrt{b}\right)^2}{y}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{x+y}\Leftrightarrow x+y\ge\left(\sqrt{a}+\sqrt{b}\right)^2\)
Đẳng thức xảy ra khi \(\frac{\frac{a}{x}}{x}=\frac{\frac{b}{y}}{y}\)
Do điều kiện là x > 0 nên sẽ khó khăn khi quy đồng và xét delta.
Áp dụng bất đẳng thức Côsi
\(A=x+\frac{5}{2x}-3\ge2\sqrt{x.\frac{5}{2x}}-3=\sqrt{10}-3\)
Dấu "=" xảy ra khi \(x=\frac{5}{2x}\text{ và }x>0\Leftrightarrow x=\sqrt{\frac{5}{2}}\)
Vậy GTNN của A là \(\sqrt{10}-3\)
\(B=\frac{x^2+\left(a+b\right)x+ab}{x}=x+\frac{ab}{x}+a+b\ge2\sqrt{x.\frac{ab}{x}}+a+b=2\sqrt{ab}+a+b\)
Dấu "=" xảy ra khi \(x=\frac{ab}{x}\text{ và }x>0\Leftrightarrow x=\sqrt{ab}\)
Vậy GTNN của B là \(2\sqrt{ab}+a+b\)
Bài 1 : \(A=\frac{2016}{x^2-2x+2017}\) đạt GTLN khi \(x^2-2x+2017\) đạt GTNN .
\(x^2-2x+2017=x^2-2x+1+2016=\left(x-1\right)^2+2016\Rightarrow GTNN\) của \(x^2-2x+2017\) là \(2016\)
\(\Rightarrow GTLN\) của \(A\) là : \(\frac{2016}{2016}=1\)
Bài 2 :
a ) Đặt \(A=\frac{2}{6x-9x^2-21}.A\) đạt \(GTNN\) Khi \(\frac{1}{A}\) đạt \(GTLN\).
Ta có : \(\frac{1}{A}=\frac{-9x^2+6x-21}{20}=-\frac{9}{20}\left(x-\frac{1}{3}\right)^2-1\le-1\)
Vậy \(Max\left(\frac{1}{A}\right)=-1\Leftrightarrow x=\frac{1}{3}\)
\(\Rightarrow Min_A=-1\Rightarrow x=\frac{1}{3}\)
b ) Đặt \(B=\left(x-1\right)\left(x-2\right)\left(x-5\right)\left(x-6\right)\)
Ta có : \(B=\left[\left(x-1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-5\right)\right]=\left(x^2-7x+6\right)\left(x^2-7x+10\right)\)
Đặt \(y=x^2-7x+8\Rightarrow B=\left(y+2\right)\left(y-2\right)=y^2-4\ge-4\)
\(Min_B=-4\) khi và chỉ khi \(x^2-7x+8=0\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{array}\right.\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
I) Đk: x > 0 và x \(\ne\)9
\(D=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(D=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
=> \(\frac{1}{D}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
Để 1/D nguyên <=> \(\frac{2}{\sqrt{x}+1}\in Z\)
<=> \(2⋮\left(\sqrt{x}+1\right)\) <=> \(\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Do \(x>0\) => \(\sqrt{x}+1>1\) => \(\sqrt{x}+1=2\)
<=> \(\sqrt{x}=1\) <=> x = 1 (tm)
\(E=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Với x\(\ge\)0; ta có:
\(E=\frac{8}{9}\) <=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+2\)
<=> \(2x-4\sqrt{x}-\sqrt{x}+2=0\)
<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
e) Ta có: \(E=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\ge0\forall x\in R\) (vì \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\))
Dấu "=" xảy ra<=> x = 0
Vậy MinE = 0 <=> x = 0
Lại có: \(\frac{1}{E}=\frac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}-1+\frac{1}{\sqrt{x}}\right)\ge\frac{3}{4}\left(2\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1\right)\)(bđt cosi)
=> \(\frac{1}{E}\ge\frac{3}{2}.\left(2-1\right)=\frac{3}{2}\)=> \(E\le\frac{2}{3}\)
Dấu "=" xảy ra<=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\) <=> x = 1
Vậy MaxE = 2/3 <=> x = 1
\(Q=\frac{x^3}{4\left(y+2\right)}+\frac{y^3}{4\left(x+2\right)}=\frac{x^3\left(x+2\right)}{4\left(x+2\right)\left(y+2\right)}+\frac{y^3\left(y+2\right)}{4\left(x+2\right)\left(y+2\right)}\)
\(=\frac{x^4+y^4+2x^3+2y^3}{4\left(x+2\right)\left(y+2\right)}=\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(xy+2x+2y+4\right)}\)
\(=\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(2x+2y+8\right)}=\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\)
Áp dụng bất đẳng thức AM-GM ta có :
\(x^4+y^4\ge2\sqrt{x^4y^4}=2x^2y^2\)
\(x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)
\(Q=\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\ge\frac{2x^2y^2+2xy\left(x+y\right)}{8\left(x+y+4\right)}=\frac{2xy\left(xy+x+y\right)}{8\left(x+y+4\right)}=\frac{8\left(x+y+4\right)}{8\left(x+y+4\right)}=1\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x,y>0\\x=y\\xy=4\end{cases}}\Rightarrow x=y=2\)
Vậy GTNN của Q là 1 <=> x = y = 2
Or
\(Q-1=\frac{\left(x^2-y^2\right)^2+2\left(x+y\right)\left(x^2+y^2-8\right)}{4\left(x+2\right)\left(y+2\right)}\ge0\)*đúng do \(x^2+y^2\ge2xy=8\)*
Do đó \(Q\ge1\)
Đẳng thức xảy ra khi x = y = 2
1. Áp dụng Min - cốp - ski, ta được: \(\sqrt{\frac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\frac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\frac{9}{\left(c+a\right)^2}+b^2}\)\(\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{c+a}\right)^2+\left(a+b+c\right)^2}\)\(\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)(Bunyakovsky dạng phân thức)
Đặt \(t=a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)thì ta cần chứng minh: \(\sqrt{\frac{729}{4t^2}+t^2}\ge\frac{3\sqrt{13}}{2}\Leftrightarrow\frac{729}{4t^2}+t^2\ge\frac{117}{4}\)\(\Leftrightarrow\frac{\left(t+3\right)\left(t-3\right)\left(2t+9\right)\left(2t-9\right)}{4t^2}\ge0\)*đúng bởi \(t-3\le0;t+3>0;2t+9>0;2t-9< 0;4t^2>0\)*
Đẳng thức xảy ra khi t = 3 hay a = b = c = 1
2. Ta có: \(\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}-3=\frac{\left(x^2-y^2\right)^2\left(x^4+y^4+x^2y^2\right)}{x^2y^2\left(x^2+y^2\right)^2}\ge0\)\(\Rightarrow\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge3\)
Đẳng thức xảy ra khi x = y
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...