a) Let \(\sqrt{x}=a\Rightarrow x=a^2\)

\(\Rightarrow P=\left[\dfrac{a^2+3a+2}{\left(a+2\right)\left(a-1\right)}-\dfrac{a^2+a}{a^2-1}\right]:\left(\dfrac{1}{a+1}+\dfrac{1}{a-1}\right)\)

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

b) \(\dfrac{1}{P}-\dfrac{\sqrt{x}+1}{8}\ge1\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}\ge1\)

\(\Leftrightarrow\dfrac{16\sqrt{x}-\left(\sqrt{x}+1\right)^2}{8\left(\sqrt{x}+1\right)}\ge1\)

\(\Leftrightarrow16\sqrt{x}-\left(\sqrt{x}+1\right)^2\ge8\left(\sqrt{x}+1\right)\) (cause \(8\left(\sqrt{x}+1\right)>0\))

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2\le0\Leftrightarrow x=9\)

Yes, thanks Lê Quốc Trần Anh :(( I will do better next time :)

There are 11 fractions can be generated by this method.

There are : \(\dfrac{1}{2};\dfrac{1}{3};\dfrac{1}{4};\dfrac{1}{5};\dfrac{1}{6};\dfrac{2}{3};\dfrac{2}{5};\dfrac{3}{4};\dfrac{3}{5};\dfrac{4}{5};\dfrac{5}{6}\)

Sum of a group of 12 people is \(26\cdot12=312\left(years\right)\)

When 8 new people added the sum is \(20\cdot32=640\left(years\right)\)

The Sum age of 8 new people is 640 - 312 = 328 (years)

The average age of the 8 new people is \(\dfrac{328}{8}=41\left(year\right)\)

If Janelle spends 100$/minute. Then she will spend 6000$/hour.

And she will spend \(6000\cdot24=144000\)$/day and \(144000\cdot7=1008000\)$/week.

It will take Janelle 1 week to spend one million dollars.

The perimeter of square A is \(4x\)

The perimeter of square B is \(8x\)

The perimeter of square C is \(12x\)

The perimeter of the entire figure is \(4x+8x+12x=24x\)(unit)

Using Venn.

The thread said that : 7 animals are neither dogs nor white.

Cause 15 animals are white and also dogs => 15 - 5 = 10 animals are white but not dogs.

And we have 7 animals are neither dogs nor white so 10 + 7 = 17 animals are not dogs.

There are 23 - 17 = 6 dogs at the pet store.

We call the Grace’s score on the fourth test is x.

After she had three tests, the sum of her three tests is \(89\cdot3=267\)

After the fourth test, the average score is \(\dfrac{267+x}{4}=91\)

Solve we get \(x=97\)

So Grace's score on the fourth test was 97.

The thread said that : 45% of the students at South Park High School were born at South Park Hospital.

=> So the percentage of students who weren't born at South Park Hospital is 55%.

The ratio is \(\dfrac{55\%}{45\%}=\dfrac{11}{9}\)

We have : 

Six lawns : Nine Hours.

One lawn : x Hours.

So \(x=\dfrac{9\cdot1}{6}=\dfrac{9}{6}=1,5\left(hour\right)\)

So it takes her 1,5 hour to mow each lawn.

Each triangle has an area of : \(\dfrac{s^2\cdot\sqrt{3}}{4}=\dfrac{4^2\cdot\sqrt{3}}{4}=4\sqrt{3}\)(inches).

SUM 4 triangles \(\left(4\sqrt{3}\right)\cdot4=16\sqrt{3}\left(inches\right)\)

The area of the square : \(4^2=16\left(inches\right)\)

The total area is \(16+16\sqrt{3}\left(inches\right)\)

Using Pythagore's Theorem we have : 

\(24^2+h^2=72^2\rightarrow h^2=4608\Rightarrow h=\sqrt{4608}\)

Using calculator we get a height of about \(67,88\left(meter\right)\)

Because the y value double the x value so : \(y=2x\)

So the function is equal \(2x=3x+6\)

We get x = -6.

So we get y = \(2\cdot\left(-6\right)=-12\)

At the point \(\left(-6;-12\right)\)the y-value double the x-value.

In the first month the fund gains 20% so the total fund of him after the first month is :

\(5000\cdot120\%=6000\)$

At the end of second month Arturo loses 20% of its value, so Arturo's investment worth at that time is :

\(6000\cdot80\%=4800\)$

So the answer is 4800$.

Sorry I have a mistake :

Because its perimeter is equal its area : 

So we call the width is x.

We have : \(8\cdot x=\left(8+x\right)\cdot2\)

\(\Rightarrow x=\dfrac{8}{3}\)

So the width is \(\dfrac{8}{3}=2\dfrac{2}{3}\left(feet\right)\)

How to solve this : If the perimeter of a regtangle is 46 units, then its half the perimeter is 46/2 = 23 units. The rectangle with interger side lengths and with a perimeter of 46 units that has the least area : 1*22=22 unit2. The rectangle with the same perimeter that has the greatest area is one with dimensions closet to those of a square. With interger side lengths, the closet we can get to a square (area) is 11/12=132 unit2.

So the difference is : 132 - 22 = 110 (unit).

That's my idea :))

Because its perimeter is equal its area.

So we called the width is x.

We have : \(8\cdot x=\dfrac{8+x}{2}\)

\(\Rightarrow x=\dfrac{8}{15}\left(feet\right)\)

So the width is \(\dfrac{8}{15}feet\)

The MAX difference between the largest and smallest possible areas of two rectangles is 110.

Shorten right side : \(\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{40+14\sqrt{2}}\)

\(=\sqrt[3]{2\sqrt{2}+3\left(\sqrt{2}\right)^2\cdot3+3\sqrt{2}\cdot9+27}-\sqrt[3]{2\sqrt{2}+3\left(\sqrt{2}\right)^2\cdot2+3\sqrt{2}\cdot4+8}\)

\(=\sqrt[3]{\left(\sqrt{2}+3\right)^3}-\sqrt[3]{\left(\sqrt{2}+2\right)^3}=\sqrt{2}+3-\left(\sqrt{2}+2\right)=1\)

So the expression is equal : \(\sqrt{2x^2+1}+\sqrt{x^2-3x+\dfrac{17}{2}}=1\)

\(\Leftrightarrow\sqrt{2x^2+1}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}}=1\)

The expression has no solution cause the left side \(>\sqrt{2x^2+1}\ge1\).

Using Cos's theorem:

We have : \(BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot cos\angle A\)

\(\Leftrightarrow5,042^2=4,123^2+7,145^2-2\cdot4,123\cdot7,415\cdot cos\angle A\)

\(\Rightarrow\angle A\simeq45,8^o\)

So on : \(AB^2=BC^2+AC^2-2\cdot BC\cdot AC\cdot cos\angle C\)

\(\Rightarrow\angle C\simeq32^o\)

Next : \(AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot cos\angle B\)

\(\Rightarrow\angle B\simeq102,2^o\)