ADTCSTSBN , ta được :

\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}=\frac{6}{2\left(a+b+c\right)}=\frac{3}{a+b+c}\)

\(\Rightarrow a+b=a+b+c\)\(\Rightarrow c=0\)

\(P=\frac{a+b-2019.0}{a+b+2018.0}=\frac{a+b}{a+b}=1\)

Vậy P = 1

ta co

3/a+b=3/b+c=3/c+a

=>1:3/a+b=1:2/b+c=1:1/c+a

=>a+b/3=b+c/2=c+a/1

ap dung DTSBN, ta có

a+b/3=b+c/2=c+a/1+(a+b)+(b+c)+(c+a)/3+2+1=2a+2b+2c/6=2.(a+b+c)/6=a+b+c/3

vi a+b/3=a+b+c/3

=>a+b=a+b+c

=>c=0

=>p=a+b-2019.0/a+b+2018.0

=>p=a+b/a+b

=>p=1

KL

https://olm.vn/hoi-dap/detail/221248297106.html

tham khảo nhé

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}=\frac{3+2+1}{a+b+b+c+c+a}=\frac{6}{2\left(a+b+c\right)}=\frac{3}{a+b+c}\)

\(\rightarrow a+b=a+b+c\)         \(\rightarrow c=0\)

\(\Rightarrow P=\frac{3a+3b+2019c}{a+b-2020c}=\frac{3\left(a+b\right)+2019\cdot0}{a+b-2020\cdot0}=\frac{3\left(a+b\right)}{a+b}=3\)

\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

Các cao nhân giúp với!!!!!!!!!! Thanks for all

Ta có:\(a+b+c\ne0\)vì nếu \(a+b+c=0\)thế vào giả thiết ta có:

\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\Leftrightarrow-3=1\)(vô lí)

Khi \(a+b+c\ne0\)ta có:

\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).\left(a+b+c\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{a.\left(b+c\right)}{b+c}+\frac{b.\left(c+a\right)}{c+a}+\frac{b^2}{c+a}+\frac{c.\left(a+b\right)}{a+b}+\frac{c^2}{a+b}=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)\(\Rightarrow P=0\)

Học tốt 

\(P=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(< =>P=a\left(\frac{a}{b+c}+1-1\right)+b\left(\frac{b}{a+c}+1-1\right)+c\left(\frac{c}{a+b}+1-1\right)\)
\(< =>P=a\left(\frac{a+b+c}{b+c}-1\right)+b \left(\frac{a+b+c}{a+c}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)\)
\(< =>P=\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{a+c}+\frac{c\left(a+b+c\right)}{a+b}-\left(a+b+c\right)\)

\(< = >P=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)

\(< =>P=a+b+c-a-b-c=0\)

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

ban oi mk dat cau hoi nay cac ban giup mk vs

1/2x + 3/5 . ( x- 2 ) = 3

Ta có :

\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\)

\(\Leftrightarrow\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1=\frac{3}{2}+3\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Leftrightarrow6.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Rightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{9}{2}:6=\frac{3}{4}\)

Vậy \(P=\frac{3}{4}\)