Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)

\(\Rightarrow x=-4k;y=-7k;z=3k\) (1)

Thay (1) vào A , ta được

\(A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)

\(\Rightarrow A=\dfrac{8k+\left(-7k\right)+15k}{-8k+21k+\left(-18k\right)}\)

\(\Rightarrow A=\dfrac{k[8+\left(-7\right)+15]}{k[-8+21+\left(-18\right)]}\)

\(\Rightarrow A=\dfrac{16k}{-5k}\)

\(\Rightarrow A=\dfrac{16}{5}\)

Vậy \(A=\dfrac{16}{5}\)

Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)

\(\Rightarrow x=-4k;y=-7k;z=3k\)(1)

Thay (1) vào ta có :

\(A=\dfrac{-2x+y+5z}{2x-3y-6z}=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{8k+-7k+15k}{\left(-8k\right)-\left(-27k\right)-18k}=\dfrac{k\left(8+-7+15\right)}{k\left(-8+27-18\right)}=\dfrac{16}{17}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

Ta có : \(\dfrac{x}{y}=\dfrac{3}{4};\dfrac{y}{z}=\dfrac{5}{6}\)

\(\Rightarrow4x=3y;6y=5z\)

\(\Leftrightarrow8x=6y=5z.\)

\(\Rightarrow y=\dfrac{8x}{6}=\dfrac{4}{3}x\)

Thay vào A ta có :

\(A=\dfrac{2x+3y+5z}{y+5z}=\dfrac{2x+4x+8x}{\dfrac{4}{3}x+8x}=\dfrac{3}{2}.\)

Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\left(k\ne0\right)\)

=>\(\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)

Ta có P =\(\dfrac{-2\cdot\left(-4k\right)+\left(-7k\right)+5\cdot3k}{2\cdot\left(-4k\right)-3\left(-7k\right)-6\left(3k\right)}\)=\(\dfrac{8k+\left(-7k\right)+15k}{-8k+21k-18k}\)=

\(\dfrac{k\cdot\left(8+\left(-7\right)+15\right)}{k.\left(-8+21-18\right)}=\dfrac{-16}{5}\)

Vậy P= \(\dfrac{-16}{5}\)

Theo đề ta có:

\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\)

Đặt k cho biểu thức trên

=>\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\) =k

=> \(\left[{}\begin{matrix}\dfrac{x}{-4}=k\\\dfrac{y}{-7}=k\\\dfrac{z}{3}=k\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\)

Thay \(\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\) vào biểu thức \(P=\dfrac{-2x+y+5z}{2x-3y-6z}\)

Ta được:

\(P=\dfrac{-2.\left(-4.k\right)+\left(-7.k\right)+5\left(3.k\right)}{2\left(-4.k\right)-3\left(-7.k\right)-6\left(3.k\right)}\)

=> \(P=\dfrac{8.k+\left(-7.k\right)+15.k}{-8.k+21.k-18.k}\)

=> \(P=\dfrac{k.\left(8+-7+15\right)}{k.\left(-8+21-18\right)}\)

=> P= \(-\dfrac{16}{5}\)

Vậy:....................

Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)

a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)

=> x=8

3y=18=>y=6

4z=72=>z=18

Vậy x=8 ; y=6 ; z=18

b, Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)

Câu c bạn làm tương tự nhé!

d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)

\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)

Vậy...