\(x+2< \dfrac{5}{x-2}\Leftrightarrow x+2-\dfrac{5}{x-2}< 0\Leftrightarrow\dfrac{x^2-9}{x-2}< 0\)

Case 1 : \(\left\{{}\begin{matrix}x^2-9< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2< 9\\x>2\end{matrix}\right.\)\(\Rightarrow2< x< 3\)

Case 2 : \(\left\{{}\begin{matrix}x^2-9>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2>9\\x< 2\end{matrix}\right.\)\(\Rightarrow-10\le x< -3\)

The answer is : \(\dfrac{\left[3-2+\left(-3\right)-\left(-10\right)\right].100}{10-\left(-10\right)}\%=40\%\)

3 mi = 190080 in

The answer is : \(190080:2327\pi\approx26\) (in)

LCM(1,2,3,4,5,6) = 60

So, they eat together every 60^th day

From 15/1 to 31/12, there are : 365 - 14 = 351 (days) or 366 - 14 = 352 (days)

The answer is : (301 - 61) : 60 + 1 = 5 (times)

1200 ml is : \(80\%\times\left(100\%-25\%\right)=60\%\) of the kettle

The answer is : \(1200:60\%=2000\) (ml)

The area of the circle is : \(6^2\pi=36\pi\) (in²)

The length of the side of the square is : \(6.\dfrac{2}{\sqrt{2}}=6\sqrt{2}\) (in)

The area of the square is : \(\left(6\sqrt{2}\right)^2=72\) (in²)

The answer is : \(36\pi-72\approx41\) (in²)

The number of customers who will receive a bag is :

(1950 - 150) : 150 + 1 = 13

The answer is : \(\dfrac{13.100}{2000}\%=0,65\%\)

Let a, b be the price of a shirt and a pair of socks respectively

\(\left\{{}\begin{matrix}2a+b=15\\a+2b=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+b=15\\2a+4b=24\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{15-b}{2}\\3b=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=6\\b=3\end{matrix}\right.\)

\(\Rightarrow15a+10b=15.6+10.3=120\)

So, the answer is 120

Draw \(RH\perp SJ;TI\perp SJ;PM\perp ST;PN\perp RT\). We have :

\(\dfrac{S_{RPS}}{S_{SPK}}=\dfrac{S_{RPS}:S_{SPT}}{S_{SPK}:S_{SPT}}=\dfrac{\dfrac{SP.RH:2}{SP.TI:2}}{\dfrac{SK.PM:2}{ST.PM:2}}=\dfrac{\dfrac{RH}{TI}}{\dfrac{SK}{ST}}\)

\(=\dfrac{\dfrac{2S_{RPJ}:PJ}{2S_{PJT}:PJ}}{\dfrac{1}{3}}=\dfrac{PN.RJ:PJ}{PN.JT:PJ}.3=\dfrac{RJ}{JT}.3=2.3=6\)

\(\Rightarrow S_{RPS}=7.6=42\left(units^2\right)\)

The speed of the hour hand is : \(\dfrac{360}{12}:60=0,5\) (⁰/m)

The speed of the minute hand is : \(360:60=6\) (⁰/m)

Two hands are perpendicular again when the minute hand goes further than the hour hand by : 90^o x 2 = 180⁰ or after : \(180:\left(6-0,5\right)=32\dfrac{8}{11}\)(minutes)

Then, the hands are perpendicular after : 

The number of balls that are replaced with green balls is :

(100 - 4) : 4 + 1 = 25

The number of balls that are replaced with white balls is :

(96 - 1) : 5 + 1 = 20

The number of balls that are replaced with yellow balls is :

(96 - 6) : 6 + 1 = 16

The number of green balls that are replaced with yellow balls is :

(96 - 12) : 12 + 1 = 8

The number of green balls that are replaced with white balls is :

(96 - 16) : 20 + 1 = 5

The number of white balls that are replaced with yellow balls is :

(96 - 6) : 30 + 1 = 4

The number of balls that are replaced three times is :

(96 - 36) : 60 + 1 = 2

The number of balls that are replaced at least once is :

5 + 2 + 8 + 4 + (25 - 5 - 2 - 8) + (20 - 5 - 2 - 4) + (16 - 8 - 2 - 4)

= 40

The answer is : 100 - 40 = 60 (balls)

Let a be the length of the side of the small square, then the diagonal of the small square or the diameter of the circle or the length of the side of the large square is \(\sqrt{2}a\). The answer is : \(\dfrac{a^2}{\left(\sqrt{2}a\right)^2}=\dfrac{1}{2}\)

\(\overline{ab3}⋮3\Rightarrow a+b+3⋮3\Rightarrow a+b⋮3\Rightarrow\overline{ab}⋮3\)

There are : (99 - 12) : 3 + 1 = 30 (choices) to choose \(\overline{ab}\)

The answer is : \(\dfrac{30}{999-100+1}=\dfrac{1}{30}\)

The surface area of the cube is : 50² x 6 = 15000 (inches²)

The volume of the cube is : 50³ = 125000 (inches³)

The answer is : 125000 - 15000 = 110000 

Let l, w be the length and the width of the rectangle respectively in inches \(\left(l\ge w>0\right)\)

\(l=\dfrac{3}{8}.2\left(l+w\right)\Leftrightarrow l=\dfrac{3}{4}l+\dfrac{3}{4}w\Leftrightarrow\dfrac{3}{4}w=\dfrac{1}{4}l\Leftrightarrow l=3w\)

\(lw=18.75\Leftrightarrow l.\dfrac{1}{3}l=18.75\Leftrightarrow l^2=56.25\Leftrightarrow l=7.5\)

So, the answer is 7.5 inches

Condition : \(x\ne3;-5\)

\(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\left(2x+34\right)\left(\dfrac{1}{15}-\dfrac{1}{\left(x-3\right)\left(x+5\right)}\right)=0\)

\(\Leftrightarrow\left(x+17\right).\dfrac{\left(x-3\right)\left(x+5\right)-15}{15\left(x-3\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=-1\pm\sqrt{31}\end{matrix}\right.\)(satisfied)

So, \(S=\left\{-17;-1\pm\sqrt{31}\right\}\)

\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\Leftrightarrow\left(\dfrac{x^2}{2}-\dfrac{x^2}{5}\right)+\left(\dfrac{y^2}{3}-\dfrac{y^2}{5}\right)+\left(\dfrac{z^2}{4}-\dfrac{z^2}{5}\right)=0\)\(\Leftrightarrow\dfrac{3}{10}x^2+\dfrac{2}{15}y^2+\dfrac{1}{20}z^2=0\)

\(\dfrac{3}{10}x^2,\dfrac{2}{15}y^2,\dfrac{1}{20}z^2\ge0\Rightarrow\dfrac{3}{10}x^2+\dfrac{2}{15}y^2+\dfrac{1}{20}z^2\ge0\)

The equality happens only when x = y = z = 0

So, x = y = z = 0

\(2x^2\left(x-3\right)=3x\left(x+2\right)-5\)

\(\Leftrightarrow2x^3-6x^2-\left(3x^2+6x\right)+5=0\)

\(\Leftrightarrow2x^3-9x^2-6x+5=0\)

\(\Leftrightarrow2x^3+2x^2-11x^2-11x+5x+5=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-11x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-11x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-10x-x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[2x\left(x-5\right)-\left(x-5\right)\right]\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x-5=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=5\\x=-1\end{matrix}\right.\)

So, \(S=\left\{\dfrac{1}{2};5;-1\right\}\)

The area of the square is : 9 x 9 = 81 (cm²)

a) x = 100 - 33 = 67

b) x = 100 - 88 = 12

c) x = 100 - 22 = 78

d) x = 100 - 18 = 82

The volume of the cube is : 2 x 2 x 2 = 8 (cm³)