The greatest distinct 4-digit number is: \(9876\). The least 3-digit number is: \(100\).

So their sum is: \(9876+100=9976\)

The final result is:

\(\left(100.2+1\right):3=67\) and \(\left(100.3+2\right):4=67\)

So the answer is: \(C.67\)

Convert: \(1.75kg=1750g\)

It take her to cook a 1.75 kg chicken: \(20+\left(1750:500.30\right)=125\left(minutes\right)\)

We put the numbers 1-2 ; 3-4 and 5-6 opposite each other. So then 2-3 and 4-5 will have the same side. So he will lose at least 2 points.

So the answer is: \(C:2points.\)

From 1 to 9 the sum of the individual digits are: \(1+2+3+4+5+6+7+8+9=45\)

From 10 to 19: the tens digit all have the same digit, the ones digit have the rules from 1 to 9.

So as the same with 20-99.

The number 100 has the individual digits sum: \(1+0+0=1\)

She has obtained the sum: \(\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+1\)

\(=450+450+1=901\)

So she obtained the sum: \(901\)

The total of the cards are: \(9+8+7+6+5+4+3+2+1=45\)

The total of the cards that can removed 1 card is: \(\left(45-1\right)to\left(45-9\right)\)

From this operation and the question, we see only \(C.54\) satisfy these.

So the answer is C

Because he threw 8 darts at the dartboard => The score must be even. (Odd + Odd = Even)

The most score that he could throw is: \(8.9=72\left(points\right)\)

The least score that he can throw is: \(8.3=24\left(points\right)\)

Satisfy all this operation, only \(C.42\) could be his score.

So the answer is C

\(516-\left[340+\left(-4-1\right)\right]=516-\left[340+\left(-5\right)\right]=516-335=181\)

\(2^3.x^2=32\)

<=> \(8.x^2=32\)

<=> \(x^2=32:8=4\)

We see: \(2^2=4\) => \(x=2\)

So the value of \(x\) is: \(2\)

Just after 8 is: 9

Just before 8 is: 7

So the product of just after and just before 8 is: \(9+7=16\).

So the answer is: \(16\)

Sorry for the error question.

The distance between the first flower and the last flower is: \(\left(60:2\right).\left(9-1\right)=240\left(cm\right)\)

So the answer is: \(240cm\)

The way that the car travel for 1 hour is: \(301:5=60,2\left(km\right)\)

So the distance that the car goes for 1 hour is: \(60,2\left(km\right)\)

Sorry, can I answer again?

\(2^1.2^2.2^3...2^n=2^{5050}\) => \(2^{1+2+3+...+n}=2^{5050}\)

=> \(1+2+3+...+n=5050\)

In the series, there are: \(\left(n-1\right):1+1\) numbers.

=> \(n.n+1:2=5050\)

=> \(n.n+1=5050.2=10100\)

We saw that: \(100.101=5050\) => \(n=100\)

To be certain, the problem must be solved by finding the largest number of fruits.

Because the number of pears are larger than the number of apples \(\left(3+5< 7+2\right)\)

=> The fruits that he must take out is: \(\left(7+2\right)+1=10\left(fruits\right)\)

So the answer is \(10fruits\)

\(2^1.2^2.2^3...2^n=2^{5050}\) => \(2^{1+2+3+...+n}=2^{5050}\)

=> \(1+2+3+...+n=5050\)

=> \(n=100\)

Because the segment BC is 1cm, and ABCD is constructed of 4 identical rectangles (In the figure) => The width of 4 small rectangles is: \(1:2=\dfrac{1}{2}\)(cm)

The area of 1 small rectangle is: \(1.\dfrac{1}{2}=\dfrac{1}{2}\left(cm^2\right)\)

Then the area of 4 small rectangle is: \(\dfrac{1}{2}.4=2\left(cm^2\right)\)

Then the segment AB is: \(2:1=2\left(cm\right)\)

1 paper strip has the length: \(\left(50-10\right):2=20\left(cm\right)\)

2 paper strips has the length: \(20.2=40\left(cm\right)\)

The length of the overlap is: \(56-40=16\left(cm\right)\)

So the length of the overlap is: \(16cm\)

1. There are: \(550.\dfrac{10}{11}=500\left(male\right)\) students

So there are: \(550+500=1050\left(students\right)\).

Sorry for the wrong answer, but Phan Thanh Tinh made the right comment.