\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2x+3y-5z}{10+12-15}=\dfrac{2x-3y+5z}{10-12+15}\\ \Rightarrow A=\dfrac{10+12-15}{10-12+15}=\dfrac{7}{13}\)

A=7/13

làm hộ, please

Ta có:

\(x:y:z=5:4:3\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=3k\end{matrix}\right.\)

\(\Rightarrow\frac{2x+3y-5z}{2x-3y+5z}=\frac{2.5k+3.4k-5.3k}{2.5k-3.4k+5.3k}=\frac{10k+12k-15k}{10k-12k+15k}=\frac{7k}{13k}=\frac{7}{13}\)

kho qua

Ta có: 2x + 3y + 5z = 2000

=> 4y + 3y + 3y = 2000

=> 10y = 2000

=> y = 2000 : 10 = 200

=> x = 200 x 4 : 2 = 400

Vậy x = 400

đổi hai cái đó ra phân số đi sau đó sử dụng nhân chéo sẽ ra

\(3x=4y;2y=5z\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{15};\dfrac{y}{15}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{2x}{40}=\dfrac{3y}{45}=\dfrac{5z}{30}\)

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

\(\dfrac{2x}{40}=\dfrac{3y}{45}=\dfrac{5z}{30}\)

\(=\dfrac{2x+3y-5z}{40+45-30}=\dfrac{55}{55}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.20=20\\y=1.15=15\\z=1.6=6\end{matrix}\right.\)

Tương tự

Ta có :

\(2x+3y-5z=55\)

\(3x=4y;2y=5z\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{2}=\dfrac{z}{5}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{2x+3y-5z}{2.19+3.12-2.16}=\dfrac{55}{22}=\dfrac{5}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{45}{2}\\\dfrac{y}{12}=\dfrac{5}{2}\Leftrightarrow x=30\\\dfrac{z}{16}=\dfrac{5}{2}\Leftrightarrow z=40\end{matrix}\right.\)

Vậy ..............

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{2};\dfrac{y}{5}=\dfrac{z}{4}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{4x-3y+5z}{60-30+40}=\dfrac{7}{70}=\dfrac{1}{10}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{15}{10}=\dfrac{3}{2}\\y=1\\z=\dfrac{8}{10}=\dfrac{4}{5}\end{matrix}\right.\)

Ta có: 2x=3y=>x/3=y/2 => 15x/15=y/10

4y=5z => y/5=z/4 => y/10 = z/8

Đặt x/15 =y/10 = z/8 = k

=> x = 15k , y = 10k , z = 8k

Ta có : 4(15k) - 3(10k) + 5(8k) = 7

=> 60k - 30k + 40k = 7

=> 70k = 7 => k = 10

Vậy :

x/15 = 10 => x = 10 . 15 = 150

y/10 = 10 => y = 10 . 10 =100

z/4 = 10 => z= 10 . 4 = 40

Có:

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

\(4y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{4}\)

Vì BCNN (2;5) = 10

\(\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{3.5}=\frac{y}{2.5}=\frac{x}{15}=\frac{y}{10}\)

\(\frac{y}{5}=\frac{z}{4}\Leftrightarrow\frac{y}{5.2}=\frac{z}{4.2}=\frac{y}{10}=\frac{z}{8}\)

⇒\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{4x}{60}=\frac{3y}{30}=\frac{5z}{40}=\frac{4x-3y+5z}{60-30+40}=\frac{7}{70}=\frac{1}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.\frac{1}{10}=\frac{3}{2}\\y=10.\frac{1}{10}=1\\z=8.\frac{1}{10}=\frac{4}{5}\end{matrix}\right.\)

GIÚP MÍNH VỚI MN ƠIIIIIIII!

\(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10};4y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{10}=\dfrac{z}{8}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{4z}{32}=\dfrac{2x+3y-4z}{30+30-32}=\dfrac{56}{28}=2\\ \Rightarrow\left\{{}\begin{matrix}x=30\\y=20\\z=16\end{matrix}\right.\)

ai cover với

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{2x+3y-3z}{2\cdot15+3\cdot10-3\cdot8}=\dfrac{96}{36}=\dfrac{8}{3}\)

Do đó: \(\left\{{}\begin{matrix}x=24\\y=\dfrac{80}{3}\\z=\dfrac{64}{3}\end{matrix}\right.\)