Câu hỏi của nguyễn Thị Hồng Thanh

Question

cho a+b+c+d khác 0 và $\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}$tìm giá trị của :$A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}$

Lê Minh Anh · Accepted Answer

$\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=$$\frac{d}{a+b+c}$$\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}$$\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}$Mà: $a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c$$\Rightarrow a=b=c=d$$\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}$$\Rightarrow A=1+1+1+1=4$Câu trả lời: $\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=$$\frac{d}{a+b+c}$$\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}$$\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}$Mà: $a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c$$\Rightarrow a=b=c=d$$\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}$$\Rightarrow A=1+1+1+1=4$

cô nàng dễ thương · Answer

số đo slaf4 nhe sbnbài dài lắm mìnhvhir tiện ghithế này thôiCâu trả lời: số đo slaf4 nhe sbnbài dài lắm mìnhvhir tiện ghithế này thôi

Thám Tử Lừng Danh Kudo S... · Answer

$\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}$$\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}$$\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}$Mà :$a+b+c+d=0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c$$\Rightarrow a=b=c=d$$\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}$$\Rightarrow A=1+1+1+1=4$Câu trả lời: $\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}$$\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}$$\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}$Mà :$a+b+c+d=0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c$$\Rightarrow a=b=c=d$$\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}$$\Rightarrow A=1+1+1+1=4$

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