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Theo T/C dãy tỉ số bằng nhau
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)
Tương tự ta có
\(b+c=2a\)
\(c+a=2b\)
Xét \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)
ta có:\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)do đó:
+)\(\dfrac{a+b-c}{c}=1\)
=> a+b-c=c
=> a+b=2c
=> a+b+c =3c (1)
cm tương tự ta đươc (bạn cần làm chi tiết hơn)
+)3a=a+b+c (2)
+) 3b=a+b+c(3)
từ (1);(2) và (3)=> 3a=3b=3c
=> a=b=c
=>B=\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{c}{c}\right)\left(1+\dfrac{b}{b}\right)=2.2.2=8\)
vậy ...
\(\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\Leftrightarrow\dfrac{1}{\dfrac{3}{a+b}}=\dfrac{1}{\dfrac{2}{b+c}}=\dfrac{1}{\dfrac{1}{c+a}}\Leftrightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}\)Đặt:
\(\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{a+c}{1}=t\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3t\\b+c=2t\\a+c=t\end{matrix}\right.\)
Ta có:\(a+b+b+c+c+a=3t+2t+t\Leftrightarrow2\left(a+b+c\right)=6t\Leftrightarrow a+b+c=3t\)
Nên:\(c=a+b+c-a-b=3t-3b=0\)
Thay vào \(A\) ta có:
\(A=\dfrac{a+b+3c}{a+b-2c}=\dfrac{a+b+0}{a+b-0}=\dfrac{a+b}{a+b}=1\)
áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=1\)
\(\Rightarrow\dfrac{a+b-c}{c}=1\Leftrightarrow a+b-c=c\Leftrightarrow a+b=2c\)
\(\Rightarrow\dfrac{b+c-a}{a}=1\Leftrightarrow b+c-a=a\Leftrightarrow b+c=2a\)
ta có
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{a+b}{a}\times\dfrac{c+a}{c}\times\dfrac{b+c}{b}=\dfrac{2c}{a}\times\dfrac{2b}{c}\times\dfrac{2a}{b}=8\)
\(\Rightarrow M=8\)
Lời giải \(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Khi \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\Leftrightarrow B=\dfrac{-abc}{abc}=-1\)
Khi \(a=b=c\Leftrightarrow B=\dfrac{8abc}{abc}=8\)
Lời giải:
Ta có: \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(Q+3=\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1\)
\(Q+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(Q+3=\frac{259}{a+b}+\frac{259}{b+c}+\frac{259}{a+c}=259\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\)
\(Q+3=259.15=3885\)
\(\Rightarrow Q=3885-3=3882\)