Đk:\(-1\le x\le3\) (chính là cái bài cho kia)

Nếu \(x=0\) thì \(A=\sqrt{3}\) ta sẽ chứng minh nó là GTNN của \(A\)

Tức là ta cần chứng minh 

\(\sqrt{-x^2+2x+3}+\sqrt{3}\le\sqrt{-x^2+4x+12}\)

Sau khi bình phương 2 vế rồi rút gọn ta cần chứng minh 

\(\sqrt{-3\left(x^2+2x+3\right)}\le x+3\)

Từ khi \(x+3>0\), ta cần chứng minh  

\(3\left(-x^2+2x+3\right)\le\left(x+3\right)^2\Leftrightarrow x^2\ge0\) (Đúng)

Vậy \(A_{Min}=\sqrt{3}\Leftrightarrow x=0\)

a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)

b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)

c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)

\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)

P<=2

=>x+1>0

=>x>-1

1)

\(M=\dfrac{1}{3}x^2+2x+10\)

\(=\dfrac{1}{3}.\left(x^2+6x+30\right)\)

\(=\dfrac{1}{3}\left(x^2+2.x.3+9\right)+7\)

\(=\dfrac{1}{3}.\left(x+3\right)^2+7\) \(\ge\) 7 với \(\forall\) x

=> M luôn dương

=> đpcm

2)

a) \(2x-x^2-15\)

\(=-\left(x^2-2x+15\right)\)

\(=-\left(x^2-2x+1\right)-14\)

\(=-\left(x-1\right)^2-14\) \(\le-14\) với \(\forall\) x

=> \(2x-x^2-15\) luôn âm

=> đpcm

b) \(-5-\left(x-1\right)\left(x+2\right)\)

\(=-5-x^2-2x+x+2\)

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

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

\(=-\left(x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)

\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\) với \(\forall\) x

=> \(-5-\left(x-1\right)\left(x+2\right)\) luôn âm

=> đpcm

\(M=\dfrac{1}{3}x^2+2x+10=\dfrac{1}{3}\left(x^2+6x+9\right)+7\)

\(=\dfrac{1}{3}\left(x+3\right)^2+7\)

Ta có:

\(\dfrac{1}{3}\left(x+3\right)^2\ge\forall x\Rightarrow\dfrac{1}{3}\left(x+3\right)^2+7>0\)

=>đpcm

\(2,a,2x-x^2-15\)

\(=-\left(x^2-2x+1\right)-14\)

\(=-\left(x-1\right)^2-14\)

Ta có:

\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-14< 0\)

=> đpcm

\(b,-5-\left(x-1\right)\left(x+2\right)\)

\(=-5-\left(x^2+x-2\right)\)

\(=-5-x^2-x+2\)

\(=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)

\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\)

Ta có:

\(-\left(x+\dfrac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x+\dfrac{1}{2}\right)-\dfrac{11}{4}< 0\)=> đpcm

a: DKXĐ: x<>1; x<>-1

b: \(A=\dfrac{x^2+2x+1+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+7-x^2+x-3x+3}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

-\(x+3+\sqrt{x^2-6x+9}\)

\(=x+3+\left|x\right|-6x+9\)

\(x< 0\)

\(--->x+3-x-6x+9\)

\(=\left(x-x\right)-6x+3+9\)

\(=-6x+\left(3+9\right)=-6x+12\)

\(x>0\)

\(--->3+x+x-6x+9\)

\(=\left(x+x-6x\right)+\left(3+9\right)\)

\(=\left(2x-6x\right)+12\)

\(=4x+12\)

a) A=6
b) B=1
 

a: ĐKXĐ: \(x\notin\left\{1;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

b: \(A=\dfrac{4x^2-4x+1-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{2x-1}{x-1}+\dfrac{2}{2x+1}\)

\(=\dfrac{4x\left(x-1\right)}{\left(2x+1\right)\left(x-1\right)}+\dfrac{2}{2x+1}\)

\(=\dfrac{4x+2}{2x+1}=2\)