P>1/3

=>P-1/3>0

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{3}>0\)

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

=>2 căn x-5>0

=>x>25/4

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

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

=>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`

\(\left(3x+1\right)^2=9\left(x-2\right)^2\)

\(\Leftrightarrow9x^2+6x+1=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow9x^2+6x+1=9x^2-36x+36\)

\(\Leftrightarrow9x^2+6x+1-9x^2+36x-36=0\)

\(\Leftrightarrow42x-35=0\)

\(\Leftrightarrow42x=35\)

\(\Leftrightarrow x=\dfrac{35}{42}=\dfrac{5}{6}\)

Vậy: \(S=\left\{\dfrac{5}{6}\right\}\)

\(\)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

Lời giải:
$117=(2y+1)^2-x^2=(2y+1-x)(2y+1+x)$

Vì $x,y$ nguyên nên $2y+1-x, 2y+1+x$ nguyên. Do đó ta có bảng sau:

Để A là số nguyên thì \(2x-1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{1;0;3;-2\right\}\)

bn ghi bắng công thức đi

\(\Leftrightarrow a\cdot\dfrac{13}{15}=\dfrac{28}{13}:2=\dfrac{14}{13}\)

=>\(a=\dfrac{14}{13}:\dfrac{13}{15}=\dfrac{210}{169}\)