已知a-b=3,a c=-5,求代数式ac-bc a^2-ab的值
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![已知a-b=3,a c=-5,求代数式ac-bc a^2-ab的值](/uploads/image/f/4211471-47-1.jpg?t=%E5%B7%B2%E7%9F%A5a-b%3D3%2Ca+c%3D-5%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8Fac-bc+a%5E2-ab%E7%9A%84%E5%80%BC)
∵3b=a+2c∴(3b)²=(a+2c)²9b²=a²+4ac+4c²∴a²-9b²+4c²=a²-(a&s
ac-bc+a^2-ab=c(a-b)+a(a-b)=(a-b)(a+c)=3*(-5)=-15
(b+a+c)(b-a-c)=-3ac,且b²=ac,b^2-(a+c)^2=-3b^4b^2-(a+c)^2=0(2b+a+c)(2b-a-c)=02b-a-c=02b+a+c=0(she
a-c=a-b+b-c=3-5=8a^2+b^2+c^2-ab-bc-ac=1/2{(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)}=1/2{(a-b)^2+(a-
ab=(a+b)/3,bc=(b+c)/4,ac=(a+c)/51/3=(a+b)/ab=1/a+1/b,1/4=1/b+1/c,1/5=1/a+1/c(1/a+1/b)+(1/b+1/c)+(1/a
(a-b)2+(b-c)2+(a-b+b-c)2=a2+b2-2ab+b2+c2-2bc+a2+b2-2ac=2(a2+b2+c2-ab-bc-ac)=3*3+5*5+(3+5)*(3+5)=9+25
ab/(a+b)=1/3(a+b)/ab=3则a/ab+b/ab=31/b+1/a=3同理1/c+1/b=41/c+1/a=5相加2(1/a+1/b+1/c)=121/a+1/b+1/c=6(ab+b
ac-bc+a·a-ab=c(a-b)+a(a-b)=(a+c)(a-b)a-b=3b+c=-5两式相加,a+c=-2所以原式=-2*3=-6
ac-bc+a2-ab=c(a-b)+a(a-b)=(c+a)(a-b)=-5*3=-15
(a+b+c)²=25(a+b)²+c²+2c(a+b)=25a²+b²+c²+2ab+2bc+2ac=253+2(ab+bc+ac)=25
已知的分别倒数后1/a+1/b=31/b+1/c=41/a+1/c=5三式相加除以2得:1/a+1/b+1/c=6abc/(ab+bc+ac)=1/(1/c+1/b+1/a)=1/6
ab/(a+b)=1/3取倒数(a+b)/ab=3a/ab+b/ab=31/b+1/a=3同理1/b+1/c=41/a+1/c=5相加2(1/a+1/b+1/c)=121/a+1/b+1/c=6通分(
因为集合的互异性所以a不等于0b不等于0c不等于+1-1(1)a+b=aca+2b=ac^2相除c=(a+2b)/(a+b)根据一式得c=(a+b)/a解得b=0舍(2)a+b=ac^2a+2b=ac
a-b=3,b+c=-5a-b+b+c=3-5=-2ac-bc+a^2-ab=c(a-b)+a(a-b)=3c+3a=3(a+c)=-2×3=-6
ab=(a+b)/3,所以3ab=a+b,所以3=1/a+1/b(1)bc=(b+c)/4,所以4bc=b+c,所以4=1/b+1/c(2)ac=(a+c)/5,所以5ac=a+c,所以5=1/a+1
若a+b=ac,a+2b=ac^2,则b=ac-a,2b=ac^2-a两式相比得,c+1=2,c=1则b=0,矛盾所以a+b=ac^2,a+2b=ac,得c=-1/2.
(a+b+c)²=a²+b²+c²+2(ab+bc+ac)3²=6+2(ab+bc+ac)ab+bc+ac=1.5
1、a+b=ac>>>c=-1/2检验,可以.所以,c=-1/2
解,原式=(ab/2c+bc/2a)+(ac/2b+ab/2c)+(bc/2a+ac/2b)》2*根号下(ab/2c*bc/2a)+2*根号下(ac/2b*ab/2c)+2*根号下(bc/2a*ac/
3/a=4/b=5/c所以a=3c/5,b=4c/5所以(ab-bc-ac)/(a^2+b^2+c^2)=(12c^2/25-4c^2/5-3c^2/5)/(9c^2/25+16c^2/25+c^2)