Terms & Expressions Overview
A brief quiz on all parts of the course book "Terms and Expressions".
A brief quiz on all parts of the course book "Terms and Expressions".
Fichier Détails
| Cartes-fiches | 12 |
|---|---|
| Utilisateurs | 19 |
| Langue | English |
| Catégorie | Mathématiques |
| Niveau | Collège |
| Crée / Actualisé | 06.11.2023 / 09.11.2023 |
| Lien de web |
https://card2brain.ch/box/20231106_terms_expressions_overview
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If we evaluate the expression \(T(x) = x^3+1\) for \(x=2 \), what do we get?
\(T(2) =\)
9
Wieso dürfen wir beim Doppelbruch \(\frac{\frac{p}{3}}{\frac{3}{4}}\)nicht folgendes tun?
\(\frac{\frac{p}{3}\cdot3}{\frac{3}{4}\cdot4}\)
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.
With which factors do we have to expand each fraction to obtain a common denominator?
\(\frac{1}{x+1}+\frac{2}{y}+\frac{1}{x}\)
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 \(\frac{y(x+1)}{y(x+1)}\).
Is the following calculation correct?
\(\frac{a}{x}+\frac{b}{y} = \frac{ay+bx}{xy}\)
Yes!
We have to expand the fractions to the LCM of the denominators. Here, \(LCM(x,y)=xy\) .
Then, \(\frac{a}{x}+\frac{b}{y} =\frac{a}{x}\cdot\frac{y}{y}+\frac{b}{y}\cdot\frac{x}{x}= \frac{ay}{xy}+\frac{bx}{xy} = \frac{ay+bx}{xy}\).
Is the following calculation correct?
\(\frac{3x-1}{3y} = \frac{x-1}{y}\)
No!
-1 was discriminated and was not divided by 3, as it should have been.
Is this calculation correct?
\(\frac{1-3x}{3x-1} = -1\)
Yes, it is correct:
\(\frac{1-3x}{3x-1} = \frac{(1-3x)}{(-1)(1-3x)} = \frac{1}{-1} =-1\)
What is the \(LCM(2^3\cdot3^3, 2^2\cdot6^2)\)?
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:
\(LCM(2^3\cdot3^3, 2^2\cdot6^2)=LCM(2^3\cdot3^3, 2^4\cdot3^2) = 2^4\cdot3^3\)
What is the \(LCM(a^2-b^2, a+b, a)\)?
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:
\(LCM(a^2-b^2, a+b, a) = LCM((a-b)(a+b),(a+b), a) =a(a-b)(a+b)\)
What is the LCM(21, 10, 4)?
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:
\(LCM(21,10,4) =LCM(3\cdot7,2\cdot5,2^2)= 2^2\cdot3\cdot5\cdot7 =420\)
How can we factorize \(x^2 -4xy -32y^2\) ?
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: \((x+4y)(x-8y) = x^2+4xy -8xy-32y^2 = x^2-4xy -32y^2\).
Simplify this expression:
\(\frac{12a-6ab}{3a}\)
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).
We need to know in which order operations (e.g., additions, divisions, raising to powers, ...) are performed.
1. What is the order of operations?
2. If we thus follow this order, what is the solution of expanding and simplifying the following expression?
\(3a(a-2b+ (-2)^2)\)
1. Brackets before powers before "Punkt" (multiplications, divisions) before "Strich" (additions, subtractions). Then, from left to right.
2. \(3a^2 -6ab+12a\)
