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bradynek
Pe: 2014-03-22T11:32:02+02:00 In: In:

Ecuatii de gradul II

Am de facut urmatoarele ecuatii:

x-(2x-3) + (3x-4)=7
〖(x+2)〗^2= 〖(2x+1)〗^2-〖3x〗^2
2x- 3/2= x/3 -4
(x+1)/3-(x-2)/2 =1+ (5x-10)/4
3(2x-1)=5(x+2)
(3x-5)/4 – (x-7)/6= 1+ (x-5)/3

^2= totul la a2-a

(x+1)/3=(x+1) supra 3
(x-2)/2= (x-2) supra 2
(5x-10)/4= (5x-10) supra 4

(3x-5)/4=(3x-5) supra 4
(x-7)/6= (x-7) supra 6
(x-5)/3= (x-5) supra 3

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  1. tata2008
    maestru (V)

    bradynek wrote: Am de facut urmatoarele ecuatii:

    x-(2x-3) + (3x-4)=7
    〖(x+2)〗^2= 〖(2x+1)〗^2-〖3x〗^2
    2x- 3/2= x/3 -4
    (x+1)/3-(x-2)/2 =1+ (5x-10)/4
    3(2x-1)=5(x+2)
    (3x-5)/4 – (x-7)/6= 1+ (x-5)/3

    ^2= totul la a2-a

    (x+1)/3=(x+1) supra 3
    (x-2)/2= (x-2) supra 2
    (5x-10)/4= (5x-10) supra 4

    (3x-5)/4=(3x-5) supra 4
    (x-7)/6= (x-7) supra 6
    (x-5)/3= (x-5) supra 3


    \rm{\bl\\ x-(2x-3) + (3x-4)=7 \rightarrow \|x-2x+3+3x+4=7\\ 2x=7-4-3\\ x=0 / 2 =0;
    \rm{\bl\\ 2x- 3/2= x/3 -4 \rightarrow \|2x - x/3 = 3/2 -4 / \cdot6\\ 12x-2x = 9-24\\ 10x = - 15\\ x = - 1,5
    \rm{\bl\\ (x+1)/3 - (x-2)/2 = 1+ (5x-10)/4 / \cdot12\\ \|4x+4 -6x+\not{12} = \not{12}+15x-30\\ 4+30 = 15x-4x+6x\\ 34 = 17x\\ x = 34:17 = 2 ;
    \rm{\bl\\ 3(2x-1)=5(x+2) \rightarrow 6x-3 = 5x +10 \\ deci 6x-5x=10+3 adica x= 13 ;
    \rm{\bl\\ \frac{3x-5}{4} - \frac{x-7}{6 } = 1+ \frac{x-5}{3} / \cdot 12\\ \|9x-15-2x-14 = 12+4x-20\\ 3x = 21\\ x = 7 ;

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