Question

Efectuați calculele : doar punctele c și d
Va rog , dau coroana

Answer 1

Răspuns:

7. c)

$\frac{2x^{2}+5x+3}{3x^{2}+7x+4 } .\frac{3x^{2}+10x+8}{2x^{2}+9x+9 }=\frac{(x+1)(2x+3)}{(x+1)(3x+4} .\frac{(3x+4)(x+2)}{(2x+3)(x+3)} =\frac{x+2}{x+3}$

7. d)

$\frac{x^{2}+8x+12}{x^{2}+4x+4 } :\frac{x^{2}+3x-18}{x^{2}-4}=\frac{(x+2)(x+6)}{(x+2)^{2} } .\frac{(x+2)(x-2)}{(x-3)(x+6)} =\frac{x-2}{x-3}$

8.c)

$\frac{x^{2}+4x}{x^{2}-6x+9} :\frac{x^{2}+9x+20}{x^{2}-7x+12}.\frac{2x+10}{3x-12} =\frac{x(x+4)}{(x-3)^{2} } .\frac{(x-3)(x-4)}{(x+4)(x+5)}.\frac{2(x+5)}{3(x-4)} =\frac{2x}{3(x-3)}$

8.d)

$\frac{4x^{2}-x-3}{2x^{2}+3x-5} .\frac{3x^{2}+10x+8}{2x^{2}+9x+9}:\frac{4x^{2}+11x+6}{2x^{2}+11x+15} =\frac{(x-1)(4x+3)}{(x-1)(2x+5)} } .\frac{(3x+4)(x+2)}{(2x+3)(x+3)}.\frac{(x+3)(2x+5)}{(x+2)(4x+3)} =\frac{3x+4}{2x+3}$

Explicație pas cu pas:

7.c)

2x²+5x+3=(x²+2x+1)+(x²+2x+1)+(x+1)=2(x+1)²+(x+1)=(x+1)[2(x+1)+1}=(x+1)(2x+3)

3x²+7x+4=(x²+2x+1)+(x²+2x+1)+(x²+2x+1)+(x+1)=3(x+1)²+(x+1)=(x+1)[(3(x+1)+1]=(x+1)(3x+4)

3x²+10x+8=3x²+4x+6x+8=x(3x+4)+2(3x+4)=(3x+4)(x+2)

2x²+9x+9=x²+6x+9+x²+3x=(x+3)²+x(x+3)=(x+3)(x+3+x)=(x+3)(2x+3)

7.d)

2x²+8x+12=(x²+4x+4)+4x+8=(x+2)²+4(x+2)=(x+2)(x+2+4)=(x+2)(x+6)

x²+4x+4=(x+2)²

x²+3x-18=x²-9+3x-9=(x+3)(x-3)+3(x-3)=(x-3)(x+3+3)=(x-3)(x+6)

x²-4=(x+2)(x-2)

8.c)

x²+4x=x(x+4)

x²-6x+9=(x-3)²

x²+9x+20=x²+8x+16+x+4=(x+4)²+(x+4)=(x+4)(x+4+1)=(x+4)(x+5)

x²-7x+12=x²-6x+9-x+3=(x-3)²-(x-3)=(x-3)(x-4)

2x+10=2(x+5)

3x-12=3(x-4)

8.d)

4x²-x-3=3x²+3+x²-x=3(x²-1)+x(x-1)=3(x+1)(x-1)+x(x-1)=(x-1)[3(x+1)+1]=(x-1)(x+4)

2x²+3x-5=2x²-2+3x-3=2(x²-1)+3(x-1)=2(x+1)(x-1)+3(x-1)=(x-1)[2(x+1)+3]=(x-1)(2x+5)

3x²+10x+8=3x²+4x+6x+8=x(3x+4)+2(3x+4)=(3x+4)(x+2)

2x²+9x+9=x²+6x+9+x²+3x=(x+3)²+x(x+3)=(x+3)(x+3+x)=(x+3)(2x+3)

4x²+11x+6=x²+4x+4+3x²+6x+x+2=(x+2)²+3x(x+2)+(x+2)=(x+2)(x+2+3x+1)=(x+2)(4x+3)

2x²+11x+15=x²+6x+9+x²+3x+2x+6=(x+3)²+x(x+3)+2(x+3)=(x+3)(x+3+x+2)=(x+3)(2x+5)