The answer: \(979+\dfrac{1}{11}\cdot979=1068\) cubic centimeters

\(\left(3^{75}\right)\cdot\left(2^{113}\right)=\left(3^{72}\right)\cdot3^3\cdot\left(...2\right)=...1\cdot...7\cdot...2=...4\)

We have:

n² + 11n + 39 = (n+9)(n+2) + 21

We see: (n+9) - (n+2) = 7\(⋮7\) so we'll have two cases.

1st case:

(n+9) and (n+2) disivided by 7

=> (n+9)(n+2) disivided by 49

but 21 not disivided by 49 => n2 + 11n + 39 not disivided by 49

2nd case:

(n+9) and (n+2) not disivided by 7

=> (n+9)(n+2) not disivided by 7

But 21 disivided by 7

=> (n+9)(n+2) + 21 not disivided by 7

So n²+11n+39 not disivided by 7

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

=> With all x \(\left(x-1\right)\left(mx+2\right)=0\)

Simple, after number is the sum of odd number in order (1, 3, 5, 7, 9) to the previous term.

=> 6th number of row is: 17 + 9 = 26

7th of row is: 26 + 11 = 37

So the 8th of row is 37+13 = 50

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

\(\Leftrightarrow3x^2+x=2x+2\Leftrightarrow3x^2=x+2\)

\(\Leftrightarrow x\left(3x-1\right)=2\)

- x = 1 => 3x-1 = 2 [satisfactory]

- x = 2 => 3x - 1 = 1 [unsatisfactory]

- x = -1 => 3x - 1 = -4 [unsatisfactory]

- x = -2 => 3x - 1 = -7 [unsatisfactory]

So x = 1

\(\dfrac{1}{\dfrac{1}{\dfrac{1}{n}+\dfrac{1}{3}}}+\dfrac{1}{\dfrac{1}{\dfrac{1}{n}+\dfrac{1}{3}}}=\dfrac{1}{n}+\dfrac{1}{3}+\dfrac{1}{n}+\dfrac{1}{3}=2\left(\dfrac{1}{n}+\dfrac{1}{3}\right)=\dfrac{2}{n}+\dfrac{2}{3}=\dfrac{5}{12}\Leftrightarrow\dfrac{2}{n}=-\dfrac{1}{4}\Leftrightarrow n=-8\)

There are \(\dfrac{157-21}{2}+1=69\) [odd numbers] between 20 and 158

There are: 35.12 = 420 [months] in 35 years

a,

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

Based on AM-GM, we have:

\(\left\{{}\begin{matrix}2004x^4+\dfrac{1}{4}\cdot2004x^2=2004x^4+501x^2\ge\left|2004x^3\right|\ge0\\1002x^2+1002\ge\left|2004x\right|\end{matrix}\right.\)

=> Left side = 0 when when x = 0

So x = 0

x² + 2xy + 7(x+y) + 2y² + 10 = 0

<=> (x² + 2xy + y²) + 7(x+y) + y² +10 = 0 

<=> (x+y)² + 7(x+y) + y² + 10 = 0(*)

P = x+y+3 <=> x+y = P-3

=> (*) = (P-3)² + 7(P-3) + y² + 10 = 0

=> P² - 6P + 9 + 7P - 21 + y² + 10 = 0

=> P² + P - 2 = -y² \(\le0\forall y\)

=> P² + P - 2 \(\le0\)

=> (P+2)(P-1) \(\le0\)

=> P+2 and P-1 diferent from sign

=> -2 < P < 1

So maximum of P is 1, minimum of P is -2

The answer is: \(\dfrac{1}{\dfrac{1}{2+\dfrac{1}{4}}}=\dfrac{1}{\dfrac{1}{\dfrac{9}{4}}}=\dfrac{1}{\dfrac{4}{9}}=\dfrac{9}{4}\)

The answer: \(\dfrac{6}{20}=\dfrac{3}{10}\)

The answer: \(\dfrac{25}{100}=\dfrac{1}{4}\)

\(\left\{{}\begin{matrix}x\left(y+z\right)=32\\y\left(x+z\right)=27\\z\left(x+y\right)=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy+xz=32\\xy+yz=27\\xz+yz=35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}xz-yz=5\\xz-xy=8\\yz-xy=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z\left(x-y\right)=5\\x\left(z-y\right)=8\\y\left(z-x\right)=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z\left(x+y-\left(x-y\right)\right)=30\\x\left(y+z-\left(z-y\right)\right)=24\\y\left(x+z-\left(z-x\right)\right)=24\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2yz=30\\2zy=24\\2xz=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}yz=15\\zy=12\\xz=12\end{matrix}\right.\Rightarrow yzzyxz=15.12.12=2160\Leftrightarrow\left(xyz\right)^2=2160\Leftrightarrow xyz=\pm12\sqrt{15}\)

We have:

\(x^2+x=x\left(x+1\right)⋮2\) is the product of two consecutive integers, so divisible by 2

\(\Rightarrow A=x^2+x+6⋮2\)

But A is a prime number and we only have an even prime number is 2

So A = 2

=> \(x^2+x+6=2\)

\(\Rightarrow x\in\varnothing\)

So there isn't any integer x satisfy the above condition

Refer at:

http://diendan.congdongcviet.com/threads/t1366::cach-kiem-tra-nhanh-n-co-phai-la-so-nguyen-to.cpp

Because this number is four-digit palindrome, we can tell this number is \(\overline{abba}\)

Because this number is greatest. We test a = 9

=> \(\overline{9bb9}⋮7and8\Leftrightarrow\overline{9bb9}⋮56\) because GCD of 7 and 8 is 1

We test b = 1 to 9 but not satisfy

Then, we test a = 8

=> \(\overline{8bb8}⋮56\)

=> b = 0

So this number to find is 8008

From 1 to 32 has the numbers as 2ⁿ is: 1;2;4;8;16;32

=> 32! has factor is: 2⁰.2¹.2².2³.2⁴.2⁵ = 2^0+1+2+3+4+5 = 2¹⁵

So n = 15