Câu hỏi của Có Anh Đây

Question

Cho ba số a, b, c khác 0 thỏa mãn $\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}.$Tính giá trị của biểu thức $M=\frac{ab+bc+ca}{a^2+b^2+c^2}.$

Đinh quang hiệp · Accepted Answer

$\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{a}{a+b}\cdot b=\frac{c}{b+c}\cdot b$$\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\Rightarrow ab+ac=ac+bc\Rightarrow ab=bc\Rightarrow a=c\left(1\right)$$\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{b}{a+b}\cdot a=\frac{c}{a+c}\cdot a$$\Rightarrow\frac{b}{a+b}=\frac{c}{a+c}\Rightarrow b\left(a+c\right)=c\left(a+b\right)\Rightarrow ab+bc=ac+bc\Rightarrow ab=ac\Rightarrow b=c\left(2\right)$$\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{b}{b+c}\cdot c=\frac{a}{a+c}\cdot c$$\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Rightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow a=b\left(3\right)$từ $\left(1\right)\left(2\right)\left(3\right)\Rightarrow a=b=c$$\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$Câu trả lời: $\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{a}{a+b}\cdot b=\frac{c}{b+c}\cdot b$$\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\Rightarrow ab+ac=ac+bc\Rightarrow ab=bc\Rightarrow a=c\left(1\right)$$\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{b}{a+b}\cdot a=\frac{c}{a+c}\cdot a$$\Rightarrow\frac{b}{a+b}=\frac{c}{a+c}\Rightarrow b\left(a+c\right)=c\left(a+b\right)\Rightarrow ab+bc=ac+bc\Rightarrow ab=ac\Rightarrow b=c\left(2\right)$$\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{b}{b+c}\cdot c=\frac{a}{a+c}\cdot c$$\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Rightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow a=b\left(3\right)$từ $\left(1\right)\left(2\right)\left(3\right)\Rightarrow a=b=c$$\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$

❤♚ℳℴℴทℛℴƴຮ♚❤ · Answer

Ta có:$\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}$$\iff$$\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}$$\iff$ $ac+bc=ab+ac=bc+ba$+)$ac+bc=ab+ac$ $\implies$$bc=ab$$\implies$ $c=a\left(1\right)$+)$ab+ac=bc+ba$$\implies$ $ac=bc$$\implies$ $a=b\left(2\right)$Từ $\left(1\right);\left(2\right)$$\implies$ $a=b=c$$\implies$ $M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$Vậy $M=1$Câu trả lời: Ta có:$\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}$$\iff$$\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}$$\iff$ $ac+bc=ab+ac=bc+ba$+)$ac+bc=ab+ac$ $\implies$$bc=ab$$\implies$ $c=a\left(1\right)$+)$ab+ac=bc+ba$$\implies$ $ac=bc$$\implies$ $a=b\left(2\right)$Từ $\left(1\right);\left(2\right)$$\implies$ $a=b=c$$\implies$ $M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$Vậy $M=1$

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