Bonsoir,
Question 19
x - y = 3
3x - y = 5
On soustrait la 1ère équation de la seconde.
(3x - y) - (x - y) = 5 - 3
3x - y - x + y = 2
2x = 2
x = 1.
On remplace x par 1 dans la 1ère équation :
1 - y = 3
-y = 3 - 1
-y = 2
y = -2
Donc : x = 1 et y = -2
y = x
2x + 4y - 2 = 0
Dans la 2ème équation, on remplace y par x :
2x + 4x - 2 = 0
6x - 2 = 0
6x = 2
x = 2/6
x = 1/3
y = x
y = 1/3
Donc : x = 1/3 et y = 1/3.
Question 18
[tex]\dfrac{a}{a+b}+\dfrac{b}{a-b}=\dfrac{a(a-b)+b(a+b)}{(a+b)(a-b)}\\\\=\dfrac{a^2-ab+ba+b^2}{a^2-b^2}\\\\=\dfrac{a^2+b^2}{a^2-b^2}[/tex]
[tex]\dfrac{3}{2xy^2}-\dfrac{5}{x^2y}-\dfrac{7}{xy}=\dfrac{3x}{2x^2y^2}-\dfrac{10y}{2x^2y^2}-\dfrac{14xy}{2x^2y^2}\\\\\\=\dfrac{3x-10y-14xy}{2x^2y^2}[/tex]
[tex]\dfrac{2a}{4a-4b}\times\dfrac{b-a}{a^2}=\dfrac{2a}{4(a-b)}\times\dfrac{b-a}{a^2}\\\\=\dfrac{a}{2(a-b)}\times\dfrac{b-a}{a^2}\\\\=\dfrac{1}{2(a-b)}\times\dfrac{b-a}{a}\\\\=\dfrac{1}{2(a-b)}\times\dfrac{-(a-b)}{a}\\\\=\dfrac{1}{2}\times(\dfrac{-1}{a})\\\\=-\dfrac{1}{2a}[/tex]
[tex]\dfrac{3a^2-6ab+3b^2}{6a^4}\times\dfrac{2a^3}{5a-5b}=\dfrac{3(a^2-2ab+b^2)}{6a^4}\times\dfrac{2a^3}{5(a-b)}\\\\=\dfrac{3(a-b)^2}{6a^4}\times\dfrac{2a^3}{5(a-b)}\\\\=\dfrac{(a-b)^2}{a^4}\times\dfrac{a^3}{5(a-b)}\\\\=\dfrac{(a-b)^2}{a}\times\dfrac{1}{5(a-b)}\\\\=\dfrac{a-b}{a}\times\dfrac{1}{5}[/tex]
[tex]=\dfrac{a-b}{5a}[/tex]
[tex]\dfrac{2a^2-18}{b-a}\times\dfrac{b}{a-3}\times\dfrac{a^2-b^2}{a+3}=\dfrac{2(a^2-9)}{b-a}\times\dfrac{b}{a-3}\times\dfrac{(a-b)(a+b)}{a+3}\\\\\\=\dfrac{2(a+3)(a-3)}{b-a}\times\dfrac{b}{a-3}\times\dfrac{(a-b)(a+b)}{a+3}\\\\\\=\dfrac{2(a-3)}{b-a}\times\dfrac{b}{a-3}\times(a-b)(a+b)[/tex]
[tex]=\dfrac{2}{b-a}\times b\times(a-b)(a+b)\\\\=\dfrac{-2}{-(b-a)}\times b\times(a-b)(a+b)\\\\=\dfrac{-2}{a-b}\times b\times(a-b)(a+b)\\\\=(-2)\times b\times(a+b)\\\\=-2b(a+b)[/tex]
