# Bài 1

* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương

* Với \(x,y>0\) áp dụng (1) ta có

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\left(\sqrt{x}\right)^2}+\dfrac{1}{\left(\sqrt{y}\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\)

Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)

* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)

Áp dụng (2) với x , y > 0 ta có

\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)

* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)

\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)

Dấu "=" xra khi \(x=y=4\)

Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)

\(-1\le sin\left(x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le2sin\left(x+\dfrac{\pi}{3}\right)\le2\)

\(\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(sin\left(x+\dfrac{\pi}{3}\right)=1\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)

\(y_{max}=5\) khi \(sin\left(x+\dfrac{\pi}{3}\right)=-1\Rightarrow x=-\dfrac{5\pi}{6}+k2\pi\)

Lời giải:
Vì $\sin (x+\frac{\pi}{3})\in [-1;1]$

$\Rightarrow y=-2\sin (x+\frac{\pi}{3})+3\in [1;5]$
Vậy $y_{\min}=1$ và $y_{\max}=5$

Ta có \(-1\le\sin\left(x-\dfrac{\pi}{3}\right)\le1\Leftrightarrow1\le-2\sin\left(x-\dfrac{\pi}{3}\right)-3\le5\)

Vậy \(y_{min}=1\) khi \(\sin\left(x-\dfrac{\pi}{3}\right)=1\)

\(y_{max}=5\) khi \(\sin\left(x-\dfrac{\pi}{3}\right)=-1\)

\(a.\dfrac{2x-1}{x-1}+\dfrac{x}{x^2-3x+2}=\dfrac{6x-2}{x-2}\left(x\ne2;x\ne1\right)\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-2\right)+x}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(6x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow2x^2-4x-x+2+x=6x^2-6x-2x+2\)

\(\Leftrightarrow2x^2-5x+2=6x^2-8x+2\)

\(\Leftrightarrow4x^2-3x=0\)

\(\Leftrightarrow x\left(4x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{3}{4}\left(TM\right)\end{matrix}\right.\)

KL........

\(b.A=\sqrt{x^2-x+1\dfrac{1}{4}}-2016=\sqrt{x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+1}-2016=\sqrt{\left(x-\dfrac{1}{2}\right)^2+1}-2016\ge1-2016=-2015\)

\(\Rightarrow A_{Min}=-2015."="\Leftrightarrow x=\dfrac{1}{2}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{2xy}\ge\dfrac{4}{1^2}+\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}\ge4+2=6\)

Dấu "=" xảy ra <=> x = y = 0,5

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)\)

\(=\left(x^2+3x+1\right)^2-1\ge-1\) với moi x

Dấu "=" xảy ra <=> x²+3x+1=0

<=>\(\left(x+\frac{3}{2}\right)^2-\frac{5}{4}=0< =>\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=0\)

\(< =>\left(x+\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x+\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=0\)

<=>..... (x có 2 nghiệm)

Vậy Min của...=-1 khi.............