What do you mean  a gives 14 and a remainder r when divided by 18 ? I don't understand much about your question ?

Moreover, remainder not remaider :))

Lê Quốc Trần Anh :)) Oops =='' Have a mistake, so the numbers of pairs are still 500 pairs with the formula this : \(\dfrac{\left(\dfrac{1000-1}{1}+1\right)}{2}=500\left(pairs\right)\).

Thanks :))

We can see that, the positive intergers which can added to make 1000 is in range :

\(0\le n\le1000\)

We can know that , each individual pair can be added to make 1000 such as (0;1000) , (1;999) , (2;998) , ....

So, the number of pairs which can be added to make 1000 is : \(\dfrac{\left(\dfrac{1000-0}{1}+1\right)}{2}=500,5\left(pairs\right)\)

So we have one number is an excess number, so the pairs which can be added to make 1000 are 500 pairs.

Dao Trong Luan you misunderstood the thread : 20203 not 20223.

:))

a)  We have 2 ways to compact :

C1 : \(P=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+x+1}{x^4-1}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)\left[x^4\left(x^2+1\right)+x^2+1\right]}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

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

C2 : Multiply numerator and denominator with x-1 we have :

\(P=\dfrac{\left(x-1\right)\left(x^7+x^6+x^5+x^4+x^3+x^2+x+1\right)}{\left(x-1\right)\left(x^4-1\right)}\)

\(=\dfrac{x^8-1}{\left(x-1\right)\left(x^4-1\right)}=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{\left(x-1\right)\left(x^4-1\right)}=\dfrac{x^4+1}{x-1}\)

b) When x = 2 we have \(P=\dfrac{2^4+1}{2-1}=\dfrac{17}{1}=17\)

So when x = 2 -> P = 17.

Use Viet system we immediately see that x,y,z are the solutions of the expression :

\(t^3-10t^2+31t-30=0\Leftrightarrow\left[{}\begin{matrix}t=5\\t=2\\t=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5;y=2z=3\\x=5;y=3;z=2\\x=2;y=5;z=3\\x=2;y=3;z=5\end{matrix}\right.\)and \(\left[{}\begin{matrix}x=3;y=2;z=5\\x=3;y=5;z=2\end{matrix}\right.\)

So we have six pairs of solutions : \(\left(5;2;3\right),\left(5;3;2\right),\left(2;5;3\right),\left(2;3;5\right),\left(3;2;5\right),\left(3;5;2\right)\)

We have : \(\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{99}=\dfrac{a}{b}\)

\(=\left(\dfrac{1}{50}+\dfrac{1}{99}\right)+\left(\dfrac{1}{51}+\dfrac{1}{98}\right)+...+\left(\dfrac{1}{74}+\dfrac{1}{75}\right)=\dfrac{a}{b}\)

\(=\dfrac{149}{50\cdot99}+\dfrac{149}{51\cdot98}+...+\dfrac{149}{74\cdot75}=\dfrac{a}{b}\)

So we have \(\dfrac{a}{b}\) kind \(\dfrac{149k}{50\cdot51\cdot...\cdot99}\)\(\Leftrightarrow\dfrac{a}{b}=\dfrac{149k}{50\cdot51\cdot...\cdot999}\Rightarrow a⋮149\left(a=149k\right)\)

Done.

Ahaaa this can be express this :

\(B=\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{96}=\dfrac{a}{b}\)

\(B=\left(\dfrac{1}{1}+\dfrac{1}{96}\right)+\left(\dfrac{1}{2}+\dfrac{1}{95}\right)+...+\left(\dfrac{1}{48}+\dfrac{1}{49}\right)\)

\(B=\dfrac{97}{96}+\dfrac{97}{190}+...+\dfrac{97}{2352}\)

We have that \(B=\dfrac{a}{b}=\dfrac{97}{96}+\dfrac{97}{190}+...+\dfrac{97}{2352}\)

Right here we have \(a\) always have kind \(97k\) which satisfy the common denominator.

\(a=97k\Rightarrow a⋮97\)

Done !

Let \(A=1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1991}{1993}\)

\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)

\(A=\dfrac{1}{1}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\Rightarrow x=1992\)

So x = 1992.

Let \(A=1^2+2^2+3^2+...+98^2\)

\(A=1\cdot1+2\cdot2+3\cdot3+...+98\cdot98\)

\(A=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+...+98\cdot\left(99-1\right)\)

\(A=\left(1\cdot2+2\cdot3+...+98\cdot99\right)-\left(1+2+...+98\right)\)

...

Let \(A=\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+...+\dfrac{1}{101}\)

We have \(A< \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)

Cause A always \(>0\), but \(A< 1\Rightarrow A\notin N\)

We have \(\dfrac{23}{7}=3+\dfrac{2}{7}=3+\dfrac{1}{\dfrac{7}{2}}=3+\dfrac{1}{3+\dfrac{1}{2}}=3+\dfrac{1}{3+\dfrac{1}{1+1}}\)

So (a;b;c)=(3;3;1).

\(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot17}\right)\)

\(\Leftrightarrow C=\dfrac{5}{11}+\dfrac{5}{22}-\dfrac{148}{187}\)

\(\Rightarrow C=-\dfrac{41}{374}\)

Call a is side of a square.

Give example a = 4. We have area S = a.a = 4.4 = 16.

Increase 20% to side we have S' = \(120\%a\cdot120\%a=23,04\)

Its area increases : \(\dfrac{23,04}{16}\cdot100\%=144\%-100\%=44\%\)

Choose D.

We have the formula of volume : \(a\cdot a\cdot a\)

Give example a = 5 we have : \(V=a^3=5^3=125\)

Increase 50% side we have : \(V'=\left(150\%\cdot a\right)^3=7,5^3=421,875\)

\(\%V_{increase}=\dfrac{\dfrac{421}{875}}{125}\cdot100\%=337,5\%\)

So the volume increase 337,5-100=237,5%.

Choose C.

I wonder if I use calculator it might gonna be :

\(\sum\limits^{100}_{x=1}\left(\dfrac{1}{x}\right)=5,187377518\)

So \(A\notin N\)

Call the bottom is a, height is h.

We have : \(\dfrac{a\cdot h}{2}=100\%\)

\(\Leftrightarrow\dfrac{120\%a\cdot80\%h}{2}=96\%\)

So the area decreases by 4%

Choose C.

\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)(1)

\(\Leftrightarrow\dfrac{xy-12}{3y}=\dfrac{1}{5}\)

\(\Leftrightarrow5xy-60=3y\)

\(\Leftrightarrow5xy-3y=60\)

\(\Leftrightarrow y\cdot\left(5x-3\right)=60\)

\(\Rightarrow y=\dfrac{60}{5x-3}\)(2).

(1),(2) => \(\dfrac{x}{3}-\dfrac{4}{\dfrac{60}{5x-3}}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{20x-12}{60}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{4\cdot\left(5x-3\right)}{60}=\dfrac{1}{5}\Leftrightarrow\dfrac{x}{3}-\dfrac{5x-3}{15}=\dfrac{1}{5}\Leftrightarrow\dfrac{5x}{15}-\dfrac{5x-3}{15}=\dfrac{1}{5}\Leftrightarrow\dfrac{3}{15}=\dfrac{1}{5}\) . 

From here we know that x has lots of value but just (x;y) = (1;30),(3;5) satisfy thread.

\(\dfrac{a}{2}+\dfrac{b}{3}=\dfrac{a+b}{2+3}\)(1)

\(\Leftrightarrow\dfrac{3a+2b}{6}=\dfrac{a+b}{5}\)

\(\Rightarrow15a+10b=6a+6b\)

\(\Rightarrow9a+4b=0\)

\(\Rightarrow a=-\dfrac{4b}{9}\)(2)

(1),(2) => \(-\dfrac{4b}{18}+\dfrac{b}{3}=\dfrac{-\dfrac{4b}{9}+b}{2+3}\)

Solve this expression we have b = 0.

So from here we have : \(\dfrac{a}{2}+0=\dfrac{a+0}{5}\Rightarrow a=0\)

So (a;b) = (0;0)

Hey I want to fix the line that :

=> BH = BC

I want to fix to : BH = HC.

Sorry for my error.