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cho hai số , b thỏa mãn a + 3b = 0 tính giá trị biểu thức M = \(\frac{2a+b}{a-b}-\frac{2a-b}{a+2b}\)
Thay a=-3b vào M
\(DK.a\ne0;b\ne0\)
\(M_b=\frac{2a+b}{a-b}-\frac{2a-b}{a+2b}=\frac{-6b+b}{-3b-b}-\frac{-6b-b}{-3b+2b}=\frac{5}{4}-\frac{-7}{-1}=-\frac{23}{4}\)
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
\(\frac{2a}{b}-\frac{2b}{a}=3\)
\(\Leftrightarrow\frac{2a^2-2b^2}{ab}=3\)
\(\Leftrightarrow2a^2-2b^2=3ab\)
\(\Leftrightarrow a=\frac{2a^2-2b^2}{3b}\)
khi đó \(S=\frac{a-b}{a+b}=\frac{\frac{2a^2-2b^2}{3b}-b}{\frac{2a^2-2b^2}{3b}+b}=\frac{2a^2-2b^2-3b^2}{\frac{3b}{\frac{2a^2-2b^2+3b^2}{3b}}}=\frac{2a^2-5b^2}{3b}.\frac{3b}{2a^2+b^2}=\frac{2a^2-5b^2}{2a^2+b^2}\)
\(=\frac{2a^2+b^2-6b^2}{2a^2+b^2}=1-\frac{6b^2}{2a^2+b^2}\)
mk chịu....đề hơi kì
cho hai số a,b thỏa mãn a+3b=0. Tính giá trị biểu thức M= \(\frac{2a+b}{a-b}\)-\(\frac{2a-b}{a+2b}\)
M=\(\frac{2a+b}{a-b}-\frac{2a-b}{a+2b}\)
M=\(\frac{2a+6b-5b}{a+3b-4b}-\frac{2a+6b-7b}{a+3b-1b}\)
M=\(\frac{2\left(a+3b\right)-5b}{\left(a+3b\right)-4b}-\frac{2\left(a+3b\right)-7b}{\left(a+3b\right)-1b}\)
M=\(\frac{-5b}{-4b}-\frac{-7b}{-1b}\)
M=\(\frac{5}{4}-\frac{7}{1}\)
M=\(-5\frac{3}{4}\)
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
