Question

Ex 7 please Și coroana

Answer 1

[tex]a)\frac{x^2+9x+20}{2x+8}\cdot\frac{4}{x^2-25}=\\ x^2+9x+20=x^2+4x+5x+20=x(x+4)+5(x+4)=\\ =(x+4)(x+5)\\ x^2-25=(x-5)(x+5)\\ Revenind:\frac{(x+4)(x+5)}{2(x+4)}\cdot\frac{4}{(x-5)(x+5)}=\boxed{\frac{2}{x-5}}\\ \\ b)\frac{x^2-10x+21}{2x^2-6x}\cdot\frac{4x^2-36}{x^2-14x+49}=\\ x^2-10x+21=x^2-3x-7x+21=x(x-3)-7(x-3)=\\ =(x-3)(x-7)\\ 2x^2-6x=2x(x-3)\\ 4x^2-36=4(x^2-9)=4(x-3)(x+3)\\ x^2-14x+49=(x-7)^2\\ Revenind:\frac{(x-3)(x-7)}{2x(x-3)}\cdot \frac{4(x-3)(x+3)}{(x-7)^2}=\frac{2(x-3)(x+3)}{x(x-7)}\\ [/tex]
[tex]c)\frac{x^2+10x+25}{x^2+6x+9}\cdot \frac{x^2-x-12}{x^2+2x-15}=\\ x^2+10x+25=(x+5)^2\\ x^2+6x+9=(x+3)^2\\ x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=\\ =(x+3)(x-4)\\ x^2+2x-15=x^2+5x-3x-15=x(x+5)-3(x+5)=\\ =(x+5)(x-3)\\ Revenind:\frac{(x+5)^2}{(x+3)^2}\cdot \frac{(x+3)(x-4)}{(x+5)(x-3)}=\frac{(x+5)(x-4)}{(x+3)(x-3)}\\ d)\frac{x^2-2x-8}{x^2+x-20}\cdot \frac{x^2+8x+15}{x^2+8x+12}=\\ x^2-2x-8=x^2-4x+2x-8=x(x-4)+2(x-4)=\\ =(x-4)(x+2)\\ x^2+x-20=x^2+5x-4x-20=x(x+5)-4(x+4)=\\ =(x+5)(x-4)\\ [/tex]
[tex]x^2+8x+15=x^2+3x+5x+15=x(x+3)+5(x+3)=\\ =(x+3)(x+5)\\ x^2+8x+12=x^2+6x+2x+12=x(x+6)+2(x+6)=\\ =(x+6)(x+2)\\ Revenind:\frac{(x-4)(x+2)}{(x+5)(x-4)}\cdot \frac{(x+3)(x+5)}{(x+6)(x+2)}=\frac{x+3}{x+6} [/tex]