\(a^3+3a^2+5=5^b\Leftrightarrow a^2\left(a+3\right)+5=5^b\Leftrightarrow a^2.5^c+5=5^b\)

\(a\ge1\Rightarrow a^3+3a^2+5\ge1^3+3.1^2+5=9\Rightarrow5^b\ge9\Rightarrow b\ge2\)

Moreover : \(a+3\ge4\Rightarrow5^c\ge4\Rightarrow c\ge1\)

So, we can divide both sides by 5 : \(a^2.5^{c-1}+1=5^{b-1}\)

Since \(b-1\ge1;c-1\ge0\), in modulo 10, we have :

\(5^{b-1}\equiv5\Rightarrow a^2.5^{c-1}\equiv4\)

If \(c-1\ge1\) : \(5^{c-1}\equiv5\Rightarrow a^2.5^{c-1}\equiv0\) or \(a^2.5^{c-1}\equiv5\) (unsatisfied)

\(\Rightarrow c-1=0\Rightarrow c=1\Rightarrow a+3=5\Rightarrow a=2\Rightarrow5^b=25\Rightarrow b=2\)

Hence, a = 2 ; b = 2 ; c = 1

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

\(=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\le4\)

The equality happens only when : \(2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)

Condition : \(x^2+8x-20=x^2+10x-2x-20=x\left(x+10\right)-2\left(x+10\right)\)

\(=\left(x-2\right)\left(x+10\right)\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2\ne0\\x+10\ne0\end{matrix}\right.\)\(\Rightarrow x\ne2;-10\)

Case 1 : \(x-2>0\Leftrightarrow x>2\)

\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\dfrac{x}{x+10}+12x-3\)

Case 2 : \(x-2< 0\Leftrightarrow x< 2\left(x\ne-10\right)\)

\(A=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\dfrac{-x}{x+10}+12x-3\)

So, \(A=\dfrac{x}{x+10}+12x-3\) when x > 2

\(A=-\dfrac{x}{x+10}+12x-3\) when \(x< 2;x\ne-10\)

Let a, a + 1, a + 2, ... , a + 10 be the integers. We have :

\(a+\left(a+1\right)+\left(a+2\right)+...+\left(a+10\right)=11\)

\(\Leftrightarrow11a+\dfrac{10.11}{2}=11\Leftrightarrow a+\dfrac{10}{2}=1\Leftrightarrow a=-4\)

So, the answer is -4

Let O be the center of the circle, then \(\widehat{AOC}=2\widehat{ABC}=2.30^0=60^0\)

The answer is : \(12\pi.\dfrac{60}{360}=2\pi\) (in)

We have : \(\left\{{}\begin{matrix}8=a.2^b\\27=a.3^b\end{matrix}\right.\)\(\Rightarrow\dfrac{2^b}{3^b}=\dfrac{8}{27}\Rightarrow\left(\dfrac{2}{3}\right)^b=\left(\dfrac{2}{3}\right)^3\Rightarrow b=3\Rightarrow a=1\)

So, a + b = 4

Let x, y be the numbers (x > y). We have :

\(\left\{{}\begin{matrix}x-y=6\\x^2-y^2=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=6\\\left(x-y\right)\left(x+y\right)=24\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=6\\x+y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y+6\\2y=-2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)\(\Rightarrow xy=-5\)

So, the answer is -5

You can use the first answer of the question : olm.vn/hoi-dap/question/908311.html to prove that n⁴ + 4ⁿ isn't a prime when n > 1 and deduce that n = 1

Since CHAIRS has 6 letters, the answer is : 6! = 720 (ways)

Let a (cm) be the width of the small rectangles, then 2a is their length and the perimeter of the rectangle PQRS is : 2(2a + a + 2a) = 10a

The area of each small rectangle is : 1536 : 3 = 512 (cm²)

We have : \(2a.a=512\Leftrightarrow a^2=256\Leftrightarrow a=16\)

The answer is : 16 x 10 = 160 (cm)

\(\dfrac{x^4.\left(x^2\right)^5}{x^5}=\dfrac{x^4.x^{10}}{x^5}=x^9\Rightarrow y=9\)

The difference between the largest element and the smallest element í the greatest difference. The largest element is 97 and the smallest element is 11. The answer is : 97 - 11 = 86

Let r be the radius of the pizza, then the length of the edge of the square is \(\sqrt{2}r\)

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

The area of the pizza is : \(r^2\pi\)

The area of the remaining pizza is : \(r^2\pi-2r^2=r^2\left(\pi-2\right)\)

The answer is : \(\dfrac{r^2\left(\pi-2\right).100}{r^2\pi}\%=100\left(1-\dfrac{2}{\pi}\right)\%\approx36\%\)

We have : \(\cos\widehat{ADB}=\dfrac{AD}{BD}=\dfrac{6}{10}=\dfrac{1}{2}\Rightarrow\widehat{ADB}=60^0\)

According to the propeties of rectangle, the diagonals are equal and intersect at the midpoint of each, so EA = EB = EC = ED

\(\Rightarrow\Delta AED\) isosceles at E

\(\widehat{AEB}\) is an exterior angle of \(\Delta AED\) , so \(\widehat{AEB}=2\widehat{ADB}=2.60^0=120^0\)

The distance between -1.3 and \(3\dfrac{1}{8}\) is : \(3\dfrac{1}{8}+1.3=4.425\)

The distance between -1.3 and the point is : \(4.425\times\dfrac{2}{3}=2.95\)

The answer is : \(-1.3+2.95=1.65\)

64.8 pounds = 1036.8 ounces

Since the volume of the cheese is directly proportional to the weight of the cheese, the ratio of the volumes of the cubes of cheese is :

\(\dfrac{1036.8}{0.6}=1728\)

The ratio of the edges of the cubes of cheese is : \(\sqrt[3]{1728}=12\)

The length of the cube of the second cheese is : 12 in = 1 ft

Let n be the number Jesse didn't include, then Jesse's sum is :

\(1+2+3+...+10-n=\dfrac{10.11}{2}-n=55-n\)

We have :\(1\le n\le10\Rightarrow45\le55-n\le54\Rightarrow55-n=49\Rightarrow n=6\)

So, the answer is 6

The area of the original rectangle is : 4 x 5 = 20 (cm²)

The total area of the removed squares is : 1² x 4 = 4 (cm²)

The answer is : 20 - 4 = 16 (cm²)

65 mi/h = 343200 ft/h = 4118400 in/h = 1144 in/s

The circumference of the tire is : \(25\pi\) (in)

The answer is : \(\dfrac{1144}{25\pi}\approx14.6\) (rotations)

The answer is : \(1:0.77\approx1.30\)