Toán Lớp 8: Tìm GTLN: a,A=-x^2-3x+10 b,B=-9x^2-6x-20 c,C=-9x(x+2/9) Giúp mik vs ạ!!

Question

Toán Lớp 8:  Tìm GTLN: a,A=-x^2-3x+10  b,B=-9x^2-6x-20  c,C=-9x(x+2/9) Giúp mik vs ạ!!

tuyethutanhthu · Answer

a) A=-x^2-3x+10 =-(x^2+3x-10) =-(x^2+2.x . 3/2+9/4-10-9/4) =-[(x+3/2)^2-49/4] =-(x+3/2)^2+49/4<=49/4 Dấu '=' xảy ra khi : (x+3/2)^2=0 <=> x+3/2=0 <=> x=-3/2 Vậy A_(max)=49/4 <=> x=-3/2 b) B=-9x^2-6x-20 =-(9x^2+6x+20) =-(9x^2+6x+1+19) =-[(3x+1)^2+19] =-(3x+1)^2-19<=-19 Dấu '=' xảy ra khi : (3x+1)^2=0 <=> x=-1/3 Vậy B_(max)=-19 <=> x=-1/3 c) C=-9x(x+2/9) =-9x^2-2x =-(9x^2+2x) =-(9x^2+2.3x. 1/3+1/9-1/9) =-[(3x+1/3)^2-1/9] =-(3x+1/3)^2+1/9<=1/9 Dấu '=' xảy ra khi : (3x+1/3)^2=0 <=> x=-1/9 Vậy C_(max)=1/9 <=> x=-1/9

quynhthanhnghi · Answer

a) A=-x^2-3x+10 A=-(x^2+3x-10) A=-(x^2+2.x. 3/2+9/4-49/4) A=49/4-(x+3/2)^2<=49/4 Dấu = xảy ra khi x+3/2=0 <=>x=-3/2 Vậy A_(max)=49/4<=>x=-3/2 b) B=-9x^2-6x-20 B=-(9x^2+6x+20) B=-[(3x)^2+2.3x+1+19] B=-19-(3x+1)^2<=-19 Dấu = xảy ra khi 3x+1=0 <=>x=-1/3 Vậy B_(max)=-19<=>x=-1/3 c) C=-9x(x+2/9) C=-9x^2-2x C=-(9x^2+2x) C=-(9x^2+2.3x. 1/3+1/9-1/9 C=1/9-(3x+1/3)^2 Dấu = xảy ra khi 3x+1/3=0 <=> 3x=-1/3 <=>x=-1/9 Vậy C_(max)=1/9<=>x=-1/9