Question

Ecuatii de gradul 1: DAU COROANA

Rezolvati in R:

a)3-2(x-1)=x+5 b)3(x-1)+4(x-2)=5 c)2+x{x-[2-(3-x)]}=1 d)radical2x-radical3=radical6x+3

Rezolvati in Z:

a)5x-2=7x-4 b)(x-1)(x+3)=(x+2)(x+4) c)(2x+3)la a doua=(3x-1)(3x+1)

Rezolvati in R ecuatiile:

a)2x+8 b)]4x=\frac{1}{5} c)3x-15=0 d)7x=0 e)-2x=frac1 pe 3

f)4x+8=0

Aflati solutiile reale ale ecuatiilor:

(3x+5)(x-6)=0

Answer 1

Răspuns:

a)3-2(x-1)=x+5

3-2x-2=x+5

3-2-5=2x+x

-4=3x

x=-3/4

b)3(x-1)+4(x-2)=5

3x-3+4x-8=5

7x=5+11

x=16/7=2,285714

c)2+x{x-[2-(3-x)]}=1

2+x[x-(2-3+x)]=1

2+x[x-(-1+x)]=1

2+x(x+1-x)=1

2+x=1

X=1-2=-1

d)radical2x-radical3=radical6x+3

R=Radical, se eleva tutto al quadrato

2x-2R6x+3= 6x+3

2x-2R6x=6x

-2R6x=4x/(-2)

R6x =-2x /^2

6x=4x^2

3x=2x^2

3x-2x^2=0

x(3-2x)=0, allora

x=0, oppure

3-2x=0

x=3/2

Explicație pas cu pas: