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

\(N=\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\) (1)

Ta có:

\(6a=4b=3c\Rightarrow\dfrac{6a}{12}=\dfrac{4b}{12}=\dfrac{3c}{12}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\) (2)

Thay (2) vào (1) ta có:

\(\dfrac{3.\left(2k\right)^2+6.\left(3k\right)^2-5.\left(4k\right)^2}{2.\left(2k\right)^2-4.\left(3k\right)^2+3.\left(4k\right)^2}=\dfrac{3.4.k^2+6.9.k^2-5.16.k^2}{2.4.k^2-4.9.k^2+3.16.k^2}\)

\(=\dfrac{12k^2+54k^2-80k^2}{8k^2-36k^2+48k^2}=\dfrac{k^2.\left(12+54-80\right)}{k^2.\left(8-36+48\right)}=\dfrac{-14}{20}=\dfrac{-7}{10}\)

Vậy giá trị của biểu thức N là \(\dfrac{-7}{10}\)

Chúc bạn học tốt!!!

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!!

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

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

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 \(\frac{3.\left(2t\right)^2+6.\left(3t\right)^2-5\left(4t\right)^2}{2.\left(2t\right)^2-4.\left(3t\right)^2+3\left(4t\right)^2}=\frac{3.4.t^2+6.9.t^2-5.16.t^2}{2.4.t^2-4.9.t^2+3.16.t^2}=\frac{-14t^2}{20t^2}=-\frac{7}{10}\)

Viết rõ đề bài ra đc không ạ

đấy là phân số

Bài làm:

a) đkxđ: \(2a\ne\pm b\)

Ta có: \(M=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right)\div\left(\frac{1+4a^2+4b^2}{4a^2-b^2}\right)\)

\(M=\left[\frac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right].\left(\frac{\left(2a-b\right)\left(2a+b\right)}{4a^2+4b^2+1}\right)\)

\(M=\frac{4a+2b-6b-8a+4b}{4a^2+4b^2+1}\)

\(M=\frac{-4a}{4a^2+4b^2+1}\)

b) +Nếu: \(a=\frac{1}{3}\)và \(b=2\)

Khi đó GT của M là: \(M=\frac{-4.\frac{1}{3}}{4.\frac{1}{3^2}+4.2^2+1}=-\frac{12}{157}\)

Viết rõ đề ra nhá

A = 2a-3b+5c-5b+4a-2c-3c+4b+7a = 13a-4b

k mk nha

A=-(-2a+3b-5c)+(-5b+4a-2c)-(3c-4b-7a)

   =2a-3b+5c-5b+4a-2c-3c+4b+7a

   =13a-4b.