7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36

Nên theo tính chất của dãy tỉ số bằng nhau ta có:

 \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6

 \(\Rightarrow\)x=6.5=30

     y=6.6=36

     z=6.7=42

vậy x=30,y=36,z=42

a.

$7x-2y=5x-3y$

$\Leftrightarrow 2x=-y$. Thay vào điều kiện số 2 ta có:

$-y+3y=20$

$2y=20$

$\Rightarrow y=10$. 

$x=\frac{-y}{2}=\frac{-10}{2}=-5$

b.

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$3y=4z-2y\Rightarrow 5y=4z\Rightarrow \frac{y}{4}=\frac{z}{5}$

$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{6+4+5}=\frac{45}{15}=3$

$\Rightarrow x=6.3=18; y=4.3=12; z=5.3=15$ 

c.

$3x=4y-2x$

$\Rightarrow 5x=4y\Rightarrow x=\frac{4}{5}y$

$3x=7z-4y$

$\Leftrightarrow \frac{12}{5}y=7z-4y$

$\Leftrightarrow \frac{32}{5}y=7z\Rightarrow z=\frac{32}{35}y$

Khi đó:

$x+y-2z=10$

$\frac{4}{5}y+y-2.\frac{32}{35}y=10$

$y.\frac{-1}{35}=10$

$y=-350$

$x=\frac{4}{5}y=\frac{4}{5}.(-350)=-280$

$z=\frac{32}{35}y=\frac{32}{35}.(-350)=-320$

Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\)

=>x=15k; y=20k; z=24k

\(A=\dfrac{2\cdot15k+3\cdot20k+4\cdot24k}{3\cdot15k+4\cdot20k+2\cdot24k}=\dfrac{186}{173}\)

\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=\dfrac{2x+3y+4z}{30+60+96}=\dfrac{3x+4y+2z}{45+80+48}\\ \Leftrightarrow A=\dfrac{2x+3y+4z}{3x+4y+2z}=\dfrac{186}{173}\)

\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}.\)VÀ \(2x-y+z=27\)

\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}=\frac{6x-12y}{9}\)\(=\frac{8z-6x}{4}=\frac{12y-8z}{16}\)

\(=\frac{6x-12y+8z-6x+12y-8z}{9+4+16}\)\(=\frac{0}{29}=0\)

\(\Rightarrow2x=4y\Rightarrow\frac{x}{4}=\frac{y}{2}\)

\(\Rightarrow4z=3x\Rightarrow\frac{z}{3}=\frac{x}{4}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

ÁP DỤNG TÍNH CHẤT CỦA DÃY TỈ SỐ BẰNG NHAU TA CÓ:

\(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}=\frac{2x-y+z}{8-2+3}\)\(=\frac{27}{9}=3\)

\(\frac{x}{4}=3\Rightarrow x=12\)

\(\frac{y}{2}=3\Rightarrow y=6\)

\(\frac{z}{3}=3\Rightarrow z=9\)

      VẬY X = 12, Y = 6, Z = 9

\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}\)

\(\Leftrightarrow\frac{6x-12y}{3^2}=\frac{8z-6x}{2^2}=\frac{12y-8z}{4^2}\)

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

\(\frac{6x-12y}{3^2}=\frac{8z-6x}{2^2}=\frac{12y-8z}{4^2}=\frac{6x-12y+8z-6x+12y-8z}{3^2+2^2+4^2}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{6x-12y}{3^2}=0\\\frac{8z-6x}{2^2}=0\\\frac{12y-8z}{4^2}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x=4y\\4z=3x\\3y=2z\end{cases}}\)  \(\Leftrightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

\(\Leftrightarrow\frac{2x}{8}=\frac{y}{2}=\frac{z}{3}\)

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

\(\frac{2x}{8}=\frac{y}{2}=\frac{z}{3}=\frac{2x-y+z}{8-2+3}=\frac{27}{9}=3\)

\(\Leftrightarrow\hept{\begin{cases}\frac{2x}{8}=3\\\frac{y}{2}=3\\\frac{z}{3}=3\end{cases}}\)  \(\Leftrightarrow\hept{\begin{cases}x=12\\y=6\\z=9\end{cases}}\)

Vậy  \(\left(x,y,z\right)=\left(12,6,9\right)\)

cái đây mà Toán 6 ak ????