Terms & Expressions Overview

Zwar würden so die Nenner in beiden Brüchen wegfallen, aber wir haben den Wert des Doppelbruchs verändert! Das dürfen wir natürlich nicht, wir müssen immer oben und unten mit der selben Zahl multiplizieren.

The LCM of the denominators is \(xy(x+1)\). Therefore, we expand the first fraction with \(\frac{xy}{xy}\), the second one with \(\frac{x(x+1)}{x(x+1)}\), and the third with \(\frac{y(x+1)}{y(x+1)}\).

Yes!

We have to expand the fractions to the LCM of the denominators. Here,  \(LCM(x,y)=xy\) .

Then, \(\frac{a}{x}+\frac{b}{y} =\frac{a}{x}\cdot\frac{y}{y}+\frac{b}{y}\cdot\frac{x}{x}= \frac{ay}{xy}+\frac{bx}{xy} = \frac{ay+bx}{xy}\).

No!

-1 was discriminated and was not divided by 3, as it should have been.

Yes, it is correct:

\(\frac{1-3x}{3x-1} = \frac{(1-3x)}{(-1)(1-3x)} = \frac{1}{-1} =-1\)

Careful, 6 is not a prime number! Therefore, we have to find its prime factors first:  \(6^2 =(2\cdot3)^2 = 2^2\cdot3^2\).

Thus, the LCM becomes:

\(LCM(2^3\cdot3^3, 2^2\cdot6^2)=LCM(2^3\cdot3^3, 2^4\cdot3^2) = 2^4\cdot3^3\)

First, we factor out all expressions. The brackets around (a+b) are not needed, but they help to see that the whole expression is one factor.

Then, we account for all available factors:

\(LCM(a^2-b^2, a+b, a) = LCM((a-b)(a+b),(a+b), a) =a(a-b)(a+b)\)

First, we have to find the prime factors of all numbers that we want to find the LCM of. Then, we form the product of the highest power of each prime factor that we found:

\(LCM(21,10,4) =LCM(3\cdot7,2\cdot5,2^2)= 2^2\cdot3\cdot5\cdot7 =420\)

We have to look for numbers whose sum is -4 and whose product is -32. The result is: 

\((x+4y)(x-8y)\). 

We can check if the result is correct by expanding it again: \((x+4y)(x-8y) = x^2+4xy -8xy-32y^2 = x^2-4xy -32y^2\).

Here, you should not start reducing right away, but factor out first.

This yields \(\frac{6a(2-b)}{3a}\). Now we have only factors in both numerator and denominator, we can start reducing. The result is  \(2(2-b)\) (factorized form), which can also be written as \(4-2b\) (expanded form).

1. Brackets before powers before "Punkt" (multiplications, divisions) before "Strich" (additions, subtractions). Then, from left to right. 

2.  \(3a^2 -6ab+12a\)