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Alinaioszip
a fost răspuns

ma puteti ajuta cu rezolvarea la exercitiile din poza? va rog frumos. dau coronita. :*

Ma Puteti Ajuta Cu Rezolvarea La Exercitiile Din Poza Va Rog Frumos Dau Coronita class=

Răspuns :

[tex]3).(21x^3-14x^2+7x):(-7x)+(x-1)(x+2)= \\ =-3x^2+2x-1+x^2+2x-x-2= \\ =-2x^2+3x-3[/tex]

[tex]\displaystyle 4). \frac{(x+1)^2}{3} - \frac{(x-2)(x+2)}{4} = \frac{(x-2)^2-6}{12} - \frac{5x+2}{6} \\ 4(x+1)^2-3(x-2)(x+2)=(x-2)^2-6-2(5x+2) \\ 4(x^2+2 \cdot x \cdot 1+1^2)-3(x^2-4)=x^2-2 \cdot x \cdot 2+2^2-6-10x-4 \\ 4(x^2+2x+1)-3x^2+12=x^2-4x+4-6-10x-4 \\ 4x^2+8x+4-3x^2+12=x^2-4x+4-6-10x-4 \\ 4x^2-3x^2-x^2+8x+4x+10x=4-6-4-4-12 \\ 22x=-22 \\ x=- \frac{22}{22} \\ x=-1[/tex]

[tex]5).y( \sqrt{3} - \sqrt{2} )+y(2 \sqrt{3} +3 \sqrt{2} )-4 \sqrt{3} y+4 \sqrt{2} y= \\ = \sqrt{3} y- \sqrt{2} y+2 \sqrt{3} y+3 \sqrt{2} y-4 \sqrt{3} y+4 \sqrt{2} y= \\ =- \sqrt{3} y+6 \sqrt{2} y=y(- \sqrt{3} +6 \sqrt{2} )[/tex]