\(\)áp dụng BĐT AM-GM(BÀi này ko có Max chỉ có Min)

\(=>\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}=\dfrac{2}{\sqrt{xy}}\)

\(=>\dfrac{1}{2}\ge\dfrac{2}{\sqrt{xy}}=>\sqrt{xy}\ge4\)

\(=>S=\sqrt{x}+\sqrt{y}\ge2\sqrt{4}=4\)

dấu"=" xảy ra<=>x=y=4

a: Ta có: \(\sqrt{2x-1}=4\)

\(\Leftrightarrow2x-1=16\)

\(\Leftrightarrow2x=17\)

hay \(x=\dfrac{17}{2}\)

b: Ta có: \(\sqrt{4x+4}-\sqrt{9x+9}=-6\)

\(\Leftrightarrow-\sqrt{x+1}=-6\)

\(\Leftrightarrow x+1=36\)

hay x=35

\(ĐK:x^2-4\ge0\Leftrightarrow x\le-2;x\ge2\)

Mặt khác : x ≥ - 2

Suy ra : x ≥ 2

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

\(a,2x\left(x^3-3\right)-2x^4=18\\ 2x^4-6x-2x^4=18\\ -6x=18\\ x=-3\)

\(b,9x\left(4-x\right)+\left(3x+1\right)^2=2\\ 36x-9x^2+9x^2+6x+1=2\\ 42x=2-1\\ 42x=1\\ x=\dfrac{1}{42}\)

\(a,\Leftrightarrow2x^4-3x-2x^4=18\Leftrightarrow-3x=18\Leftrightarrow x=-6\\ b,\Leftrightarrow36x-9x^2+9x^2+6x+1=2\\ \Leftrightarrow42x=1\Leftrightarrow x=\dfrac{1}{42}\)

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

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

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

\(\left|x+1\right|=3\\ \left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=2\left(loai\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

với x=-4 thì

\(\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)

\(=>P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{-2x-4}{x^2-4}\)`(x ne +-2)`

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

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

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

`|x+1| =3`

`=>[(x+1=3),(x+1=-3):}`

`=> [(x=3-1=2(ktm) ),(x=-3-1=-4(t/m)):}`

Thay `x=-4` vào `P` ta đc

`P= (-4)/(-4+2) = 2`

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

=>x=-1

`6/x+1/2=2`

`6/x=2-1/2`

`6/x=3/2`

`x=6:3/2`

`x=4`

Vậy `x=4`