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Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\) =\(\frac{a+b+c+d}{b+c+d+a+c+d+a+b+d+a+b+c}\)
Vì a+b+c+d khác 0
=> b+c+d=a+c+d=a+b+d=a+b+c
=>a=b=c=d
Khi đó:
a + b = c+d
b+c= (a+d)
c+d=a+b
d+a=b+c
=>\(\frac{a+b}{c+d}=\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Câu hỏi của Đoàn Thị Như Thảo - Toán lớp 7 - Học toán với OnlineMath
Ta có :
\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-1-1-1\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
Thay \(a+b+c=2001\)và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10};\)có :
\(A=2001.\frac{1}{10}-3\)
\(=200,1-3\)
\(=197,1\)
Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=197,1\)
Vì \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
=> \(\frac{a}{b+c}+1=\frac{b}{c+a}+1=\frac{c}{a+b}+1\)
=> \(\frac{a+b+c}{b+c}=\frac{a+b+c}{c+a}=\frac{a+b+c}{a+b}\)
Nếu a + b + c = 0
=> a + b = -c
b + c = -a
a + c = - b
Khi đó P = \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Khi a + b + c \(\ne0\)
=> \(\frac{1}{b+c}=\frac{1}{c+a}=\frac{1}{a+b}\)
=> b + c = c + a = a + b
=> a = b = c
Khi đó P = \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)
Vậy khi a + b + c = 0 => P = -3
khi a + b + c \(\ne0\)=> P = 6
Ta có; \(\frac{a+b+c}{c}=\frac{a+b}{c}+1;\frac{b+c-a}{a}=\frac{b+c}{a}-1;\frac{c+a-b}{b}=\frac{c+a}{b}-1\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)
\(\Rightarrow\frac{a+b-2c}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
\(\Rightarrow\frac{a}{c}+\frac{b}{c}-2=\frac{c}{b}+\frac{a}{b}=\frac{b}{a}+\frac{c}{a}\)
Ta có; a+b+cc =a+bc +1;b+c−aa =b+ca −1;c+a−bb =c+ab −1⇒a+bc +1=b+ca −1=c+ab −1
⇒a+b−2cc =b+ca =c+ab
⇒ac +bc −2=cb +ab =ba +ca
Ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
\(\Rightarrow\frac{a}{b+c}+1=\frac{b}{c+a}+1=\frac{c}{a+b}+1\)
\(\Rightarrow\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
TH 1 : \(a+b+c\ne0\)
Mà \(\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
\(\Rightarrow b+c=a+c=a+b\)
\(\Rightarrow a=b=c\)
Lại có : \(A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\)
\(\Rightarrow A=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}\)
\(\Rightarrow A=1+1+1\)
\(\Rightarrow A=3\)
TH 2 : \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}b+c=-a\\c+a=-b\\a+b=-c\end{cases}}\)
Lại có : \(A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\)
\(\Rightarrow A=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}\)
\(\Rightarrow A=-1+-1+-1\)
\(\Rightarrow A=-3\)
Vậy \(\orbr{\begin{cases}A=3\\A=-3\end{cases}}\)
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{b+c}=\frac{1}{2}\Rightarrow a=\frac{b+c}{2}\)
\(\frac{b}{c+a}=\frac{1}{2}\Rightarrow b=\frac{c+a}{2}\)
\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow c=\frac{a+b}{2}\)
Ta có: \(A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\)( sửa đề một chút )
\(\Rightarrow A=\frac{b+c}{\frac{b+c}{2}}+\frac{c+a}{\frac{c+a}{2}}+\frac{a+b}{\frac{a+b}{2}}\)
\(A=\frac{2.\left(b+c\right)}{b+c}+\frac{2.\left(c+a\right)}{c+a}+\frac{2.\left(a+b\right)}{a+b}\)
\(A=2+2+2\)\(\left(b+c;c+a;a+b\ne0\right)\)
\(A=6\)
Vậy \(A=6\)
Tham khảo nhé~
*Nếu \(a+b+c\ne0\)thì ta áp dụng t/c dãy tỉ số bằng nhau:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
*Nếu \(a+b+c=0\)thì \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=-1\)