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

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

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

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

xy³+4xy³-3xy³

=5xy³-3xy³ = 2xy³

tươg tự

Bài 2 : Thay zô có j kó đâu ==

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!!

a³-4a²b=2b³-5ab²

=>(a³-3a²b+3ab²-b³)-(a²b+b³-2ab²)=0

=>(a-b)³-b(a²-2ab+b²)=0

=>(a-b)²(a-2b)=0

=> a-2b=0 (vì a#b#0 bạn thiếu điều kiện nha)

=>a=2b. Thay a=2b vào bt P ta đc P=1

Giải:

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\)

\(\Rightarrow a=2k,b=3k,c=4k\)

Ta có: \(\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)

\(=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}\)

\(=\frac{2^2.k^2+3^2.k^2+2.4^2.k^2}{2^2.k^2-4.3^2.k^2+4^2.k^2}\)

\(=\frac{4.k^2+9.k^2+32.k^2}{4.k^2-36.k^2+16.k^2}\)

\(=\frac{k^2.\left(4+9+32\right)}{k^2.\left(4-36+16\right)}\)

\(=\frac{45}{-16}\)

\(A=\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)

Suy ra \(A=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}=\frac{4k^2+9k^2+2\cdot16k^2}{4k^2-4\cdot9k^2+16k^2}\)

\(=\frac{k^2\left(4+9+32\right)}{k^2\left(4-36+16\right)}=\frac{45}{-16}=-\frac{45}{16}\)

khong biet