Condition : \(x\ge0\)

We have :

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

We got this : 

Following inequality Cauchy , we have :

x² + 1 \(\ge2\sqrt{x^2.1}=2x\)

=> \(\dfrac{2x}{x^2+1}\le\dfrac{2x}{2x}=1\)

=> \(1+\dfrac{2x}{x^2+1}\le2\)

Hence : \(Max_P=2\)

<=> \(\dfrac{2x}{x^2+1}=1\)

<=> 2x = x² + 1

<=> (x - 1)² = 0

<=> x = 1

So , the anwser is A

\(C=\dfrac{8-\dfrac{8}{5}+\dfrac{8}{25}-\dfrac{8}{125}}{9-\dfrac{9}{5}+\dfrac{9}{25}-\dfrac{9}{125}}:\dfrac{161616}{151515}\)

\(C=\dfrac{8.\left(1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125}\right)}{9.\left(1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125}\right)}.\dfrac{15}{16}\)

\(C=\dfrac{8}{9}.\dfrac{15}{16}=\dfrac{1.5}{1.2}=\dfrac{5}{2}\)

\(\dfrac{2^2.3^2.4^2.5^2}{28800}=\dfrac{\left(2.3.4.5\right)^2}{28800}=\dfrac{120^2}{2.14400}=\dfrac{120^2}{2.120^2}=\dfrac{1}{2}\)

Percentage of people have black hair is :

\(100\%-12\%-25\%=63\%\)

The number of people have black hair is :

\(600.63\%=\dfrac{600.63}{100}=6.63=378\) (people)

P is a natural number 

<=> 4n + 1 \(⋮\) 2n + 3

<=> 4n + 6 - 6 + 1 \(⋮\) 2n + 3

<=> 2(2n + 3) - 5 \(⋮\) 2n + 3

<=> 5 \(⋮\) 2n + 3

=> 2n + 3 \(\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

We have a table :

Hence , n = {-1 ; -2 ; 1 ; 4}

\(⋮\) 

We have :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

.................

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\) (1)

In another case , we have :

\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

................

\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{100.101}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{99}{202}\) (2)

From (1) and (2)

=> \(\dfrac{99}{100}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{99}{202}\)

\(\dfrac{2}{7}-\left(\dfrac{3}{8}+\dfrac{9}{7}\right)\)

\(=\dfrac{2}{7}-\dfrac{3}{8}-\dfrac{9}{7}\)

\(=-\dfrac{7}{7}-\dfrac{3}{8}\)

\(=-1-\dfrac{3}{8}\)

\(=-\dfrac{11}{8}\)

3^{3m + 1} : 3^{2m - 1} = 9

3^{(3m + 1) - (2m - 1)} = 3²

<=> 3m + 1 - 2m + 1 = 2

<=> m + 2 = 2

=> m = 0

\(\left\{{}\begin{matrix}x^2y+xy^2+x+y=9\\xy+2x+2y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)+x+y=9\\xy+2x+2y=8\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=S\\xy=P\end{matrix}\right.\) , ta có :

\(\left\{{}\begin{matrix}P.S+S=9\\P+2S=8\end{matrix}\right.\)

\(\left\{{}\begin{matrix}S.\left(8-2S\right)+S=9\\P=8-2S\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2S^2+9S-9=0\\P=8-2S\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}S=\dfrac{3}{2}\\P=5\end{matrix}\right.\) hoặc \(\begin{matrix}S=3\\P=2\end{matrix}\)

Just replace it , you have done the exam 

Number of students receiving B is :

\(40.25\%=\dfrac{40.25}{100}=\dfrac{40}{4}=10\) (students)

Anwser : 10 students 

We got these things from that picture .

Because a//b 

=> \(\widehat{x}+\widehat{y}=180^0\)

\(\dfrac{x}{y}=\dfrac{1}{4}\Leftrightarrow\dfrac{x}{1}=\dfrac{y}{4}\)

Apply the same sequence properties , we have : 

\(\dfrac{x}{1}=\dfrac{y}{4}=\dfrac{x+y}{1+4}=\dfrac{180}{5}=36^0\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.36^0=36^0\\y=4.36^0=144^0\end{matrix}\right.\)

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{4}\)

\(\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{3}{4}\)

\(\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{3}{4}\)

\(\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}=\dfrac{3}{4}\)

\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}=\dfrac{3}{4}\)

=> (x + 1)(x + 4) = 4

=> x² + 5x + 4 = 4

=> x² + 5x = 0

=> x(x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Hence , x = 0 or x = -5 

\(A=\dfrac{4^2.3^{10}}{2^3.9^5}=\dfrac{\left(2^2\right)^2.3^{10}}{2^3.\left(3^2\right)^5}=\dfrac{2^4.3^{10}}{2^3.3^{10}}=\dfrac{2^4}{2^3}=2\)

Because 2 > 1

So A > B

I can do this, if I'm wrong, just point me to the wrong place ! Thank you !

We got this : 

(x + y)² \(\le\) 2(x² + y²) 

( because x² + 2xy + y² \(\le\) 2x² + 2y²

          =>  x² + y² \(\ge\) 2xy

          =>  x² + y² - 2xy  \(\ge\) 0

        <=> (x - y)² \(\ge\) 0 this is true         )

Means : (x + y)² \(\le\) 2(x² + y²) = 2.1 = 2                           (1)

=> \(\left(x+y\right)^2\le2\)

<=>  \(x+y\le\sqrt{2}\)

or \(-\left(x+y\right)\ge-\sqrt{2}\)                                           (2)

But \(\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^2+y^2}{xy}=\dfrac{1}{xy}\ge\dfrac{1}{\dfrac{\left(x^2+y^2\right)}{2}}=\dfrac{1}{\dfrac{1}{2}}=2\)

( \(xy\ge\dfrac{\left(x^2+y^2\right)}{2}\) because \(2xy\ge x^2+y^2\))

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge2\)                                                                   (3)

We have : 

\(P=\left(1+x\right)\left(1+\dfrac{1}{y}\right)+\left(1+y\right)\left(1+\dfrac{1}{x}\right)\)

\(P=1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{1}{y}\)

\(P=2+\left(\dfrac{1}{x}+\dfrac{1}{y}+x+y\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

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

Apply inequality AM-GM , we have :

\(2x+\dfrac{1}{x}\ge2.\sqrt{\dfrac{2x.1}{x}}=2\sqrt{2}\)

\(2y+\dfrac{1}{y}\ge2.\sqrt{\dfrac{2y.1}{y}}=2\sqrt{2}\)

\(-\left(x+y\right)\ge-\sqrt{2}\)     (from (2))

\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)                 (from (3))

So : 

\(P\ge2+2\sqrt{2}+2\sqrt{2}-\sqrt{2}+2\)

\(P\ge4+3\sqrt{2}\)

The equation occurs :

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1\\2x=\dfrac{1}{x}\\2y=\dfrac{1}{y}\end{matrix}\right.\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+.....+\dfrac{1}{17.18.19.20}\)

\(=\dfrac{1}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+....+\dfrac{3}{17.18.19.20}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+......+\dfrac{1}{17.18.19}-\dfrac{1}{18.19.20}\right)\)

\(=\dfrac{1}{3}.\dfrac{-1}{18.19.20}=\dfrac{-1}{3.18.19.20}\)

\(\dfrac{1.2+2.4+3.6}{2.4+4.8+6.12}=\dfrac{1.2+2.4+3.6}{2.1.2.2+2.2.2.4+2.3.2.6}\)

\(=\dfrac{1.2+2.4+3.6}{4.1.2+4.2.4+4.3.6}\)

\(=\dfrac{1.2+2.4+3.6}{4.\left(1.2+2.4+3.6\right)}=\dfrac{1}{4}\)

x²y + xy²

= x.xy + xy.y

= xy.(x + y)

= 3.4 = 12

So , x²y + xy² = 12

x² + 4x + 5

= x² + 2.2.x + 2² + 1

= (x + 2)² + 1 

We have : \(\left(x+2\right)^2\ge0\)

=> \(\left(x+2\right)^2+1\ge1\)

=> Min = 1

<=> (x + 2)² = 0

<=> x + 2 = 0

<=> x = -2 

We have this :

a³ + b³ = (a + b)(a² - ab + b²)        (From the equality constant)

<=> a³ + b³ = 3.(5 - ab) 

And : a + b = 3

=> (a + b)² = 9

=> a² + 2ab + b² = 9

but a² + b² = 5

So 2ab = 4 this equality 

=> ab = 2

Change ab = 2 into a³ + b³ = 3.(5 - ab) 

=> a³ + b³ = 3.(5 - 2) = 3.3 = 9 

So : a³ + b³ = 9

5x(x - 2000) - x + 2000 = 0

5x(x - 2000) - (x - 2000) = 0

(5x - 1)(x - 2000) = 0

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)