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bài này làm như sau:
\(\frac{ab}{abc}=\frac{bc}{bca}=\frac{ca}{cab}=\frac{ab+bc+ca}{abc+bca+cab}=\frac{10a+b+10b+a+10c+a}{100a+10b+c+100b+10c+a+100c+10a+b}\)
\(=\frac{11a+11b+11c}{111a+111b+111c}=\frac{11\left(a+b+c\right)}{111\left(a+b+c\right)}=\frac{11}{111}\)
vậy k=11/111
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a: \(P\left(x\right)=6x^3-x-1+5x^3+x^2-2x-1=11x^3+x^2-3x-2\)
b: \(Q\left(x\right)=6x^3-x-1-5x^3-x^2+2x+1=x^3-x^2+x\)
c: \(2\cdot U\left(x\right)=5x^3-2x+x^2-1-3\cdot\left(6x^3-x-1\right)\)
\(\Leftrightarrow2\cdot U\left(x\right)=5x^3+x^2-2x-1-18x^3+3x+3\)
\(\Leftrightarrow2\cdot U\left(x\right)=-13x^3+x^2+x+2\)
hay \(U\left(x\right)=-\dfrac{13}{2}x^3+\dfrac{1}{2}x^2+\dfrac{1}{2}x+1\)
\(a,P\left(x\right)=A\left(x\right)+B\left(x\right)=6x^3-x-1+5x^3-2x+x^2-1=11x^3+x^2+x-2\)
\(b,Q\left(x\right)=A\left(x\right)-B\left(x\right)=6x^3-x-1-5x^3+2x-x^2+1=x^3-x^2+x\)
c, Ta có : \(2U\left(x\right)+2A\left(x\right)=B\left(x\right)\)
hay \(2U\left(x\right)+2\left(6x^3-x-1\right)=5x^3-2x+x^2-1\)
\(\Rightarrow2U\left(x\right)+12x^3-2x-2=5x^3-2x+x^2-1\)
\(\Rightarrow2U\left(x\right)=5x^3-2x+x^2-1-12x^3+2x+2\)
\(\Rightarrow U\left(x\right)=-7x^3+x^2+1\)
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\(\frac{ab+ac}{2}\)=\(\frac{ba+bc}{3}\)=\(\frac{ca+cb}{4}\)=\(\frac{2\left(ab+ac+bc\right)}{9}\)(áp ụng tính chất dãy tỉ số bằng nhau)
*\(\frac{ab+ac}{2}\)=\(\frac{2\left(ab+ac+bc\right)}{9}\)=> 4,5(ab+ac)=2(ab+ac+bc) =>4,5ab+4,5ac=2ab+2ac+2bc=>2,5ab+2,5ac=2bc(rút gọn)
=>5(ab+ac)=4bc(1)=>1,25 (ab+ac)=bc
*\(\frac{ab+ac}{2}\)=\(\frac{ba+bc}{3}\)=\(\frac{ba+1,25ab+1,25ac}{3}\)=\(\frac{2,25ab+1,25ac}{3}\)
=>3(ab+ac)=2(2,25ba+1,25ac)=>3ab+3ac=4,5ba+2,5bc
=>0,5ac=1,5ba=>ac=3ab(2)
thay (2) vào (1) ta có 5(ab+3ab)=4bc=>5.4ab=4bc=> 5a=c (rút gọn) =>a/1=c/5(3)
Mà ac=3ab=>c=3b=>c/3=b/1 (4)
từ (3) và (4) suy ra: a/1=c/5 ;b/1=c/3=>\(\frac{a}{3}\) =\(\frac{b}{5}\) = \(\frac{c}{15}\) (đpcm)
sau có bài nào tương tự thì cứ hỏi mình nhá
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\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)
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