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Bài giải
Thay \(x=\frac{a}{m}\text{ ; }y=\frac{b}{m}\text{ ; }z=\frac{a+b}{m}\) vào \(P\) ta được :
\(P=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+m}{m}}{\frac{a+2b}{m}}=\frac{a+b}{m}\cdot\frac{m}{a+2b}=\frac{a+b}{a+2b}\)
Áp dụng :
\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{4}\cdot\frac{4}{5}=\frac{3}{5}\)
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)
\(=\frac{a+ab+1}{a+ab+1}=1\)
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 \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=197,1\)
Từ a+b+c=2010
\(\Rightarrow\)a= 2010-(b+c)
\(\Rightarrow\)b= 2010-(c+a)
\(\Rightarrow\)c= 2010-(a+b)
Thay vào A, ta được:
A=\(\frac{2010-\left(b+c\right)}{b+c}\)+ \(\frac{2010-\left(c+a\right)}{c+a}\) + \(\frac{2010-\left(a+b\right)}{a+b}\)
A= \(\frac{2010}{b+c}\)+ \(\frac{2010}{c+a}\)+\(\frac{2010}{a+b}\)- 3
A= 2010( \(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)+\(\frac{1}{a+b}\) ) -3
A= 2010. \(\frac{1}{10}\)-3
A=201-3
A= 198
Vậy A=198
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
a, \(p=\frac{x+y}{y+z}=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+b}{m}}{\frac{a+b^2}{m}}=\frac{a+b}{a+b^2}\)
\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{1}{4}+\frac{2}{4}}{\frac{2}{4}+\frac{1+2}{4}}=\frac{1+2}{1+2^2}=\frac{3}{5}\)
Hok tốt !!!!!!!!!
Có: x:y:z=2:3:5
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow xyz=2k.5k.3k=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
=> x=...
y=...
z=...
Có: VT\(\ge0\)( tự xét )
Theo bài ra lại có: VT\(\le0\)
=> VT=0
\(\Rightarrow\hept{\begin{cases}x_1p=y_1q\\.............\\x_mp=y_mq\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x_1}{y_1}=\frac{q}{p}\\...............\\\frac{x_m}{y_m}=\frac{q}{p}\end{cases}}\)
\(\Rightarrow\frac{x_1}{y_1}=\frac{x_2}{y_2}=.....=\frac{x_m}{y_m}=\frac{q}{p}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
........................................................................
những bài khác chốc về làm nốt cho
Ta có: \(\hept{\begin{cases}a+b+c=2020\\\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=2021\end{cases}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2020\cdot2021\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+a}+\frac{a}{c+a}+\frac{b}{b+c}+\frac{c}{b+c}=2020\cdot2021\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+1+1+1=2020\cdot2021\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2020\cdot2021-3=4082417\)