M=2a-5b/a-3b với a/b=3/5 giúp tui với
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`a/b=3/5=>a=3/5b`
Thay `a=3/5b` vào `[2a-5b]/[a-3b]` có:
`[2. 3/5b-5b]/[3/5b-3b]`
`=[6/5b-5b]/[3/5b-3b]`
`=[-19/5b]/[-12/5b]`
`=[-19/5]/[-12/5]=19/12`
\(\dfrac{2a-5b}{a-3b}=\dfrac{2\left(\dfrac{a}{b}\right)-5}{\left(\dfrac{a}{b}\right)-3}=\dfrac{2.\dfrac{3}{4}-5}{\dfrac{3}{4}-3}=\dfrac{14}{9}\)
a)Thay \(x=\dfrac{-2}{3}\) vào\(x^3-6x^2-9x-3\):
\(\left(\dfrac{-2}{3}\right)^3-6\left(\dfrac{-2}{3}\right)^2+9.\dfrac{2}{3}-3\)
\(=\dfrac{-8}{27}-\dfrac{8}{3}+6-3\)
\(=\dfrac{-8-72}{27}+3=\dfrac{-80}{27}+3=\dfrac{1}{27}\)
b) Ta có: \(\dfrac{a}{b}=\dfrac{3}{4}\Rightarrow a=3k;b=4k\)
\(\Rightarrow\dfrac{2a-5b}{a-3b}=\dfrac{6k-20k}{3k-12k}=\dfrac{-14k}{-9k}=\dfrac{14}{9}\)
c) Có: a-b=7\(\Rightarrow a=b+7\)
Thay vào \(\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{2b+21}{2b+21}+\dfrac{2b-7}{2b-7}\)
\(=1+1=2\)
TA CÓ\(\frac{2A-5B}{A-3B}=2\frac{A}{B}-5\) / A-3B
=\(2.\left(\frac{3}{4}\right)-5\)/ 3/4-3
=\(\frac{14}{9}\)
Ta có: 2a = 3b => \(\dfrac{a}{3}=\dfrac{b}{2}\)
Ta có: 5b = 6c => \(\dfrac{b}{6}=\dfrac{c}{5}\)
Ta có: \(\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{6}=\dfrac{c}{5}\Rightarrow\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}\)
và a + 3b - 2c = -5
Áp dụng t/c dãy tỉ số = nhau; ta có:
\(\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+3b-2c}{9+3.6-2.5}=\dfrac{-5}{17}\)
\(\dfrac{a}{9}=\dfrac{-5}{17}\) => a = -45/17
\(\dfrac{b}{6}=\dfrac{-5}{17}\) => b = -30/17
\(\dfrac{c}{5}=\dfrac{-5}{17}\) => c = -25/17
Vậy... a = -45/17
b = -30/17
c = -25/17.
Ta có:
+) \(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{18}=\dfrac{b}{12}\)
+) \(5b=6c\Rightarrow\dfrac{b}{6}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{10}\)
=> \(\dfrac{a}{18}=\dfrac{b}{12}=\dfrac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{18}=\dfrac{b}{12}=\dfrac{c}{10}\Rightarrow\dfrac{a}{18}+\dfrac{3b}{36}-\dfrac{2c}{20}=\dfrac{a+3b-2c}{18+36-20}=-\dfrac{5}{34}\)
Suy ra:
\(\dfrac{a}{18}=-\dfrac{5}{34}\Rightarrow a=-\dfrac{45}{17}\)
\(\dfrac{b}{12}=-\dfrac{5}{34}\Rightarrow b=-\dfrac{30}{7}\)
\(\dfrac{c}{10}=-\dfrac{5}{34}\Rightarrow c=-\dfrac{25}{17}\)
\(\frac{a}{b}=\frac{3}{4}\Rightarrow\frac{a}{3}=\frac{b}{4}\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k\Rightarrow a=3k;b=4k\) Thay vào \(\frac{2a-5b}{a-3b}\) ta được :
\(\frac{2a-5b}{a-3b}=\frac{2.3k-5.4k}{3k-3.4k}=\frac{6k-20k}{3k-12k}=\frac{k\left(6-20\right)}{k\left(3-12\right)}=\frac{-12}{-9}=\frac{4}{3}\)
2a-5b/a-3b =\(\frac{2\left(\frac{a}{b}\right)-5}{\frac{a}{b}-5}\) =2(3/4)-5/3/4-5
=14/9
\(2a=3b;5b=7c\Rightarrow\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{7}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14};\dfrac{b}{14}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
áp dụng dãy tỉ số bằng nhau ta có:
\(\dfrac{a3+c5-b7}{21.5+10.5-14.7}=\dfrac{30}{15}=2\)
\(\Rightarrow a=2.21=42\)
\(b=14.2=28\)
\(c=10.2=20\)
Ta có: \(\dfrac{a}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{5}\)
Đặt \(\dfrac{a}{3}=\dfrac{b}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3k\\b=5k\end{matrix}\right.\)
Ta có: \(\dfrac{2a-4b}{a-5b}\)
\(=\dfrac{2\cdot3k-4\cdot5k}{3k-5\cdot5k}=\dfrac{6k-20k}{3k-25k}\)
\(=\dfrac{-14k}{-22k}=\dfrac{7}{11}\)
tui học lớp 4 chứ có phải lớp 7 đâu.
\(M=2a-\dfrac{5b}{a}-3b\)
\(\dfrac{a}{b}=\dfrac{3}{5}\Rightarrow a=\dfrac{3}{5}b\) và \(\dfrac{b}{a}=\dfrac{5}{3}\)
\(\Rightarrow M=2.\dfrac{3}{5}b-5.\dfrac{5}{3}-3b\)
\(\Rightarrow M=\dfrac{6}{5}b-3b-\dfrac{25}{3}\)
\(\Rightarrow M=\left(\dfrac{6}{5}-3\right)b-\dfrac{25}{3}\)
\(\Rightarrow M=\dfrac{-9}{5}b-\dfrac{25}{3}\)