A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus,.

Whenis multiplied by, the numerators and denominators cancel out, and you are left with 1.

Another way to represent this question is:

In the one's column,andadd to produce a number with a two in the one's place. In the ten's column, we can see that a one must carry in order to get a digit in the hundred's place. Together, we can combine these deductions to see that the sum ofandmust be twelve (a one in the ten's place and a two in the one's place).

In the one's column:

The one carries to the ten's column.

In the ten's column:

The three goes into the answer and the one carries to the hundred's place. The final answer is 132. From this, we can see thatbecause.

Using this information, we can solve for.

You can check your answer by returning to the original addition and plugging in the values ofand.

First we want to simplify the expression:.

笨人的一种方式lify this complex fraction is to find the least common multiple of all the denominators, i.e. the least common denominator (LCD). If we find this, then we can multiply every fraction by the LCD and thereby be left with only whole numbers. This will make more sense in a little bit.

The denominators we are dealing with are 2, 3, 4, 5, and 6. We want to find the smallest multiple that these numbers have in common. First, it will help us to notice that 6 is a multiple of both 2 and 3. Thus, if we find the least common multiple of 4, 5, and 6, it will automatically be a multiple of both 2 and 3. Let's list out the first several multiples of 4, 5, and 6.

4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

The smallest multiple that 4, 5, and 6 have in common is 60. Thus, the least common multiple of 4, 5, and 6 is 60. This also means that the least common multiple of 2, 3, 4, 5, and 6 is 60. Therefore, the LCD of all the fractions is 60.

Let's think of the expression we want to simplify as one big fraction. The numerator contains the fractions 1/4, 1/3, and –1/5. The denominator of the fraction is 1/2, –1/6 and 1. Remember that if we have a fraction, we can multiply the numerator and denominator by the same number without changing the value of the fraction. In other words,x/y= (xz)/(yz). This will help us because we can multiply the numerator (which consists of 1/4, 1/3, and –1/5) by 60, and then mutiply the denominator (which consists of 1/2, –1/6, and 1) by 60, thereby ridding us of fractions in the numerator and denominator. This process is shown below:

==

=

This means thata/b= 23/80. We are told thataandbare both positive and that their greatest common factor is 1. In other words,a/bmust be the simplified form of 23/80. When a fraction is in simplest form, the greatest common factor of the numerator and denominator equals one. Since 23/80 is simplified,a= 23, andb= 80. The sum ofaandbis thus 23 + 80 = 103.

The answer is 103.

To convert from a fraction to a mixed number we must find out how many times the denominator goes into the numerator using division and the remainder becomes the new fraction.

平均,我们必须加值、除by two. To do this we need to find a common denomenator of 6. We then add and divide by 2, yielding 4.5/6. This reduces to 3/4.

This problem is solved the same way ½ + 1/3 is solved. For example, ½ + 1/3 = 3/6 + 2/6 = 5/6. Find a common denominator then convert each fraction into an equivalent fraction using that common denominator. The final step is to add the two new fractions and simplify.

Set up the conversions as fractions and solve:

To simplify exponents which are being divided, subtract the exponents on the bottom from exponents on the top. Remember that only exponents with the same bases can be simplified

x²–y²can be also expressed as (x+y)(x–y).

Therefore, the fraction now can be re-written as (x+y)(x–y)/(x+y).

This simplifies to (x–y).

Notice that theterm appears frequently. Let's try to factor that out:

Now multiply both the numerator and denominator by the conjugate of the denominator: