. Rezolvaţi în Q ecuațiile: x + 5 X- 37 . 5 - 3x + 3 4 2 6. 4 3 7x - 5 + 4 - 5x 4- 15,2; 4x + 3 1 - 6x _ 19. + 3 1- 2 5 d) 7. 3 21 e) ** x + 2 x + 4 _ x + 1 3x 11 3 5 3 15' f 4 4 4x - 5 3x + 2 2 2 2 + + + 5. 6 3 3 4 2 9x + 13 3x – 5 – 5x - 1 - 5 - 2x + 1. 30 4 30 5 3x + 7 i) x-2 5x - 1 1. x-5 6 2 12 2 2 3 1) 2x + 5+ 385 1 --2-5 x + 2 - 2 - 3x + 7 _ 2x + 5. X-6 2 g) -- h) - - + 15_6 j) 2 6 - 2x = 7 - 5(x + 1).DAU COROANA
Răspunsuri la întrebare
Răspuns:
Explicație pas cu pas:
a) (x+5)/3+(x-3)/4=7/2 Aducem la acelasi numitor ⇒[4(x+5)+3(x-3)]=6·7/12 ⇒4x+20+3x-9=42 ⇒7x+11=42 ⇒7x=31 ⇒x=31/7
b) (-x+7)/6-(5-3x)/4=2 1/3 Aducem la acelasi numitor si introducem intregii in fractie ⇒[2(-x+7)-3(5-3x)]/12=(4·7)/12 ⇒-2x+14-15+21x=28 ⇒19x-1=28 ⇒19x=29 ⇒x=29/19
c) (7x-5)/2+(4-5x)/5=15,2 Aducem la acelasi numitor ⇒
[5(7x-5)+2(4-5x)]/10=152/10 ⇒35x-25+8-20x=152 ⇒15x=152+17 ⇒15x=169 ⇒x=169/15
d) (4x+3)/7-(1-6x)/3=19/21 Aducem la acelasi numitor ⇒
[3(4x+3)-7(1-6x)]/21=19/21 ⇒12x+9-7+42x=19 ⇒56x=17 ⇒x=17/56
e) (x+2)/3-(x-4)/5=(x+1)/3-2x/15 Aducem la acelasi numitor ⇒
[5(x+2)-3(x-4)]/15=[5(x+1)-2x]/15 ⇒5x+10-3x+12=5x+5-2x ⇒2x+22=3x+5 ⇒2x-3x=5-22 ⇒-x=-17 ⇒x=17
f) 2x+5+(3x-5)/4-11/4=-2-(x-6)/2 Aducem la acelasi numitor ⇒
[4(2x+5)+3x-5-11]/4=2[-2-2(x-6)]/4 ⇒8x+20+3x-16=-4-4x+24 ⇒11x+4=-4x+20 ⇒11x+4x=20-4 ⇒15x=15 ⇒x=1
g) (4x-5)/6-(3x+2)/3-2/3=(3x+7)/4-(2x+5)/2 Aducem la acelasi numitor ⇒
[2(4x-5)-4(3x+2)-4·2]12=[3(3x+7)-6(2x+5)]/12 ⇒8x-10-12x-8-8=9x+21-12x-30 ⇒-4x-26=-3x-9 ⇒-x=26-9 ⇒-x=17 ⇒x=-17
h) (9x+13)/30-(3x-5)/4=(5x-1)/30-(2x+1)/5 Aducem la acelasi numitor ⇒
[2(9x+13)-15(3x-5)]/60=[2(5x-1)-6(2x+1)]/60 ⇒18x+26-45x+75=10x-2-12x-6 ⇒-27x+101=-2x-8 ⇒-25x=-109 ⇒x=109/25
i) (3x+7)/6+(x-2)/2=(5x-1)/12-1/2 Aducem la acelasi numitor ⇒[2(3x+7)+6(x-2)]/12=[(5x-1)-6]/12 ⇒6x+14+6x-12=5x-1-6 ⇒12x+2=5x-7 ⇒7x=-9 ⇒x=-9/7
j) (x-5)/2-(6-2x)/3=7-5(x+1) Aducem la acelasi numitor ⇒
[3(x-5)-2(6-2x)]/6=6[7-5(x+1)]/6 ⇒3x-15-12+4x=42-30x-30 ⇒
7x-27=-30x+12 ⇒7x+30x=12+27 ⇒37x=39 ⇒x=39/37