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徒山岩3263若a,b,c属于正整数R+,且a+b+c=1,则根号a+根号b+根号c的最大值为若a,b,c属于正整数R+,且a+b+c=1,则根号a+根号b+根号c的最大值为 -
湛卢唯15256494426 ______[答案] 利用基本不等式:2√ab ≤a+b (√a+√b+√c)^2 =a+b+c+2√ab+2√bc+2√ca =1+2√ab+2√bc+2√ca ≤1+[(a+b)+(b+c)+(c+a)] =1+2[a+b+c] =3 ∴√a+√b+√c≤√3

徒山岩3263设a,b,c为△ABC的三条边,化简:√(a+b - c)²+√(a - b - c)² - √(c+a - b)². -
湛卢唯15256494426 ______ 因为三角形任意两边之和大于第三边 也就是说任意两边之和-第三边>0 a+b-c>0 b+c-a>0 c+a-b>0 √(a+b-c)²+√(a-b-c)²-√(c+a-b)²=(a+b-c)+(b+c-a)+(c+a-b)=a+b+c

徒山岩3263sin(B+C)=?cos(B+C)=? -
湛卢唯15256494426 ______[答案] SIN(B+C)=sinBcosC+cosBsinC COS(B+C)=cosBcosC-sinBsinC

徒山岩326321、某反应A+B→2C,其速率方程为r=kcAcB,则此反应为双分子反应...
湛卢唯15256494426 ______[答案] (a+b+c²)=a^2+b^2+c^2+2(ab+bc+ac) >=1/2(2ab+bc+ac)+2(ab+bc+ac) =3(ab+bc+ac),a=b=c等号成立 (a+b+c²)=a^2+b^2+c^2+2(ab+bc+ac)

徒山岩3263三角形ABC,求sinA+sinB+sinC的值域. -
湛卢唯15256494426 ______[答案] sinA+sinB+sinC=sinB(1+cosC)+sinC(1+cosB) =4cos(B/2)cos(C/2)[sin(B/2)cos(C/2)+cos(B/2)sin(C/2)] =4cos(A/2)cos(B/2)cos(C/2) 比较不等式xyz≤(x^3+y^3+z^3)/3 A=B=C时,cos(A/2)=cos(B/2)=cos(C/2) sinA+sinB+sinC最大=3√3/2 sinA,sinB,sinC...

徒山岩3263若a+b+c=0,1a+1+1b+2+1c+3=0,那么(a+1)2+(b+2)2+(c+3)2=______. -
湛卢唯15256494426 ______[答案] ∵a+b+c=0, ∴(a+1)+(b+2)+(c+3)=6, 两边平方得(a+1)2+(b+2)2+(c+3)2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36, 又由 1 a+1+ 1 b+2+ 1 c+3=0去分母,得 (b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0, ∴(a+1)2+(b+2)2+(c+3)2=36. 故答案...

徒山岩3263如果(a+b+c)(a+b+c)(a+b+c)的解是 -
湛卢唯15256494426 ______[答案] 因为(a+b+c)小于0不等于0,所以不等式两边同时除以(a+b+c),不等号变号,小于号变为大于号(因为(a+b+c)

徒山岩3263在三角形ABC中 ∠B+∠C=2∠A ∠B:∠A=5:3 则∠A=?? ∠C=??? -
湛卢唯15256494426 ______ ∠B:∠A=5:3 设 ∠B=5k ∠A=3k ∠B+∠C=2∠A ∠C=2∠A-∠B=2*3k-5k=k ∠A+∠B+∠C=3k+5k+k=9k=180度 k=20度 ∠C=k=20度 ∠A=3k=3*20=60度

徒山岩3263因式分解(ab+a)+(b+1) -
湛卢唯15256494426 ______[答案] (ab+a)+(b+1) =a(b+1)+(b+1) =(b+1)(a+1)