`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:

\(\dfrac{a}{18}=-\dfrac{5}{34}\Rightarrow a=-\dfrac{45}{17}\)

\(\dfrac{b}{12}=-\dfrac{5}{34}\Rightarrow b=-\dfrac{30}{7}\)

\(\dfrac{c}{10}=-\dfrac{5}{34}\Rightarrow c=-\dfrac{25}{17}\)

Cách nhanh nhất :

Theo bài ra :

2a=3b=>2a-3b=0 (1)

5b=6c=> 5b-6c=0 (2)

a+3b-2c=-5 (3)

Từ (1) , (2) , (3) ta có hệ :

\(\frac{a}{b}=\frac{3}{4}\Rightarrow\frac{a}{3}=\frac{b}{4}\)

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

Sai rồi Đinh Đức Hùng ạ

6-20=-14 còn 3-12=-9 cơ mà

Sửa lại đi !

Chúc các bạn hok tốt nha

Nhớ k đúng cho mk

\(2a=3b;5b=7c\Rightarrow\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{7}=\dfrac{c}{5}\)

\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14};\dfrac{b}{14}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

áp dụng dãy tỉ số bằng nhau ta có:

\(\dfrac{a3+c5-b7}{21.5+10.5-14.7}=\dfrac{30}{15}=2\)

\(\Rightarrow a=2.21=42\)

\(b=14.2=28\)

\(c=10.2=20\)

Vì 5b = 7c => 7c - 5b = 0 Mà 7c - 5b = 3(a - 10) => 3(a-10) = 0 => a - 10 = 0 => a = 10 Vì 2a = 3b => 2.10 = 3b => b = \(\dfrac{20}{3}\) Vì 5b = 7c => c = \(\dfrac{5b}{7}\)= \(\dfrac{5.\dfrac{20}{3}}{7}\)= \(\dfrac{100}{21}\) Vậy a = 10 ; b = \(\dfrac{20}{3}\); c = \(\dfrac{100}{21}\)

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: \(\dfrac{2a-4b}{a-5b}\)

\(=\dfrac{2\cdot3k-4\cdot5k}{3k-5\cdot5k}=\dfrac{6k-20k}{3k-25k}\)

\(=\dfrac{-14k}{-22k}=\dfrac{7}{11}\)