1) 16х(2-х)+(4х-5)=0 2) 9у(у+6)-(3у+1)=-1

Question

1) 16х(2-х)+(4х-5)=0
2) 9у(у+6)-(3у+1)=-1

Answer 1

1) 32х-16х^2+4х-5=0;
36х-16х^2-5=0; :(-16);
х^2-36/16х+5/16=0;
х^2-9/4х+5/16=0;
х^2-2*9/8х+81/64-81/64+5/16=0;
(х-9/8)^2= 81/64-20/64= 61/64;
х-9/8= +-(61^1/2)/8;
х1=9/8-(61^1/2)/8; х2=9/8+(61^2)/8;
2) 9у^2+54у-3у-1+1=0;
9у^2+51у=0;
у(9у+51)=0;
у=0;
9у+51=0, 9у=-51, у=-51/9, у=-17/3=-5цел 2/3;
у1=0; у2=-5цел 2/3;