Algebraic identities, remarkable identities, literal calculation, quadratic equations, factorization, (a - b)², a² - 2ab + b², algebraic expressions, quadratic formulas, math exercises, equation solving, variable calculation, identity verification, algebraic manipulations, math problems, equation factorization, quadratic identity, algebraic simplification, math solutions, equation calculation, identity application, algebraic proof, math identities, quadratic expression, factorization techniques, algebraic equations, math problem solving, (a + b)², (a - b)² identity, quadratic equation solving, algebraic methods, math exercises solutions
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Abstract
This document explains literal calculation and remarkable identities, providing examples and step-by-step solutions to exercises.
Contents
Exercise 1
Exercice 2
Exercice 3
Exercise 4
Exercice 5
Exercice 6
Exercice 7
Exercice 8
Exercise 1
Exercice 2
Exercice 3
Exercise 4
Exercice 5
Exercice 6
Exercice 7
Exercice 8
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[...] / We reason here in a similar way to question by noting the remarkable identity 25x² = a², we find a = 5x = b², we find b = 9 on vérifie 2 ab = 2*5x*9 = 10x*9 = 90x Here, we notice the remarkable identity a² - b² = Exercise 3 : A product is null when at least one of the two factors is null, we therefore have : soit soit The equation admits two solutions which are and 3 A product is null when at least one of the two factors is null : soit so it The equation admits two solutions which are and A product is null when at least one of the two factors is null Let The solutions to the equation are 0 and For me here, there is an error, because if we take the remarkable identity a² + 2ab + b², we get a = 3x and b = 2 but 2ab = 2 * 3x * 2 = 12x and not 18 that's why we must always check the value of 2ab A product is null when at least one of the two factors is null, here the two factors are equal The solution to the equation is x = 2 A product is null when at least one of the two factors is null Let The two solutions to the equation are 11 and -11 Exercise 4 : It is the first figure because the area of the colored surface is broken down into the area of the large square surface (i.e. 81) minus the area of the small square i.e. [...]
[...] Literal calculation Exercise 1 : These are remarkable identities, which will have to be memorized by heart. Exercise 2 : c Here, we notice the remarkable identity a² - 2ab + b² = - b)². We first find b = 2 because b² = then since -12t = -2ab, so -12t = -2*2a = we find a = 3t. [...]
