\(6a=4b=3c\Leftrightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\hept{\begin{cases}a=2k\\b=3k\\c=4k\end{cases}}\)

xong bạn thay vô biểu thức N rút gọn là ra

Ta có:

6a = 4b = 3c

=> \(\dfrac{6a}{12}=\dfrac{4b}{12}=\dfrac{3c}{12}\)

=> \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

=> \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

Đặt \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)= k

=>\(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)

Thay \(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)vào biểu thức N ta được:

N = \(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)

N = \(\dfrac{3.4k+6.9k-5.16k}{2.4k-4.9k+3.16k}\)

N = \(\dfrac{12k+54k-80k}{8k-36k+48k}\)

N = \(\dfrac{-14k}{20k}\)

N = \(\dfrac{-7}{10}\)

6a = 4b  = 3c => \(\frac{6a}{12}=\frac{4b}{12}=\frac{3c}{12}\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=t\)

=> a = 2t ; b = 3t ; c = 4t thay vào N ta có \(\frac{3.\left(2t\right)^2+6.\left(3t\right)^2-5\left(4t\right)^2}{2.\left(2t\right)^2-4.\left(3t\right)^2+3\left(4t\right)^2}=\frac{3.4.t^2+6.9.t^2-5.16.t^2}{2.4.t^2-4.9.t^2+3.16.t^2}=\frac{-14t^2}{20t^2}=-\frac{7}{10}\)

B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)

=-5c+3a-4b-3a+4b-7c+12b+6a-15c

=6a +12b -27c

C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)

=32b+12c-5a+2c-4b-23a-17a+16c+31b+6b-3c

=-45a+65b+9c

bn chỉ cần thêm bước đặt nhân tử chung thui nhé!!

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r