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Ta có:
\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}.\)
\(\Rightarrow\frac{a+b}{3}=\frac{b+c}{2}=\frac{c+a}{1}.\)
Đặt \(\frac{a+b}{3}=\frac{b+c}{2}=\frac{c+a}{1}=k\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=2k\\c+a=1k\end{matrix}\right.\)
Có \(a+b+b+c+c+a=3k+2k+1k\)
\(\Rightarrow2a+2b+2c=\left(3+2+1\right).k\)
\(\Rightarrow2.\left(a+b+c\right)=6k\)
\(\Rightarrow a+b+c=6k:2\)
\(\Rightarrow a+b+c=3k.\)
\(\Rightarrow c=3k-a-b\)
\(\Rightarrow c=3k-3b\)
\(\Rightarrow c=0.\)
Lại có: \(P=\frac{3a+3b+2019c}{a+b-2020c}\)
\(\Rightarrow P=\frac{3a+3b+2019.0}{a+b-2020.0}\)
\(\Rightarrow P=\frac{3a+3b+0}{a+b-0}\)
\(\Rightarrow P=\frac{3a+3b}{a+b}\)
\(\Rightarrow P=\frac{3.\left(a+b\right)}{a+b}\)
\(\Rightarrow P=3.\)
Vậy \(P=3.\)
Chúc bạn học tốt!
ta có : 3/a+b=2/b+c=1/c+a=>a+b/3=b+c/2=c+a/1
=a+b-b-c/3-2=a-c/1
=>c+a=a-c=>c=0=>b=2a
thay c=0;b=2a vào M ta đc:
M=2a+3.2a+2020.0/3a+2.2a-2021.0=8a/7a=8/7
Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\\\frac{1}{c}+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}\end{cases}}\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Áp dụng t/c dtsbn:
\(\dfrac{1}{a+b}=\dfrac{2}{b+c}=\dfrac{3}{c+a}=\dfrac{1+2+3}{2\left(a+b+c\right)}=\dfrac{6}{2\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)
\(\Rightarrow\left\{{}\begin{matrix}3a+3b=a+b+c\\3b+3c=2a+2b+2c\\3a+3c=3a+3b+3c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=2a\\b=0\end{matrix}\right.\)
\(Q=\dfrac{a+2021b+c}{a+2022b+c}=\dfrac{a+2a}{a+2a}=1\)
Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
\(\frac{1}{a+b}=\frac{2}{b+c}=\frac{3}{c+a}=\frac{1+2+3}{2\left(a+b+c\right)}=\frac{3}{a+b+c}.\)
\(\Rightarrow\frac{3}{c+a}=\frac{3}{a+b+c}\Rightarrow c+a=a+b+c\Rightarrow b=0\)
\(\Rightarrow Q=\frac{a+2021b+c}{a+2022b+c}=\frac{a+c}{a+c}=1\)
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:
\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}=\frac{3+2+1}{a+b+b+c+c+a}=\frac{6}{2\left(a+b+c\right)}=\frac{3}{a+b+c}\)
\(\rightarrow a+b=a+b+c\) \(\rightarrow c=0\)
\(\Rightarrow P=\frac{3a+3b+2019c}{a+b-2020c}=\frac{3\left(a+b\right)+2019\cdot0}{a+b-2020\cdot0}=\frac{3\left(a+b\right)}{a+b}=3\)