1)

a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )

5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)

\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)

b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)

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

\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)

\(\dfrac{3x}{2}=\dfrac{4y}{3}=\dfrac{6z}{5}\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{5}{6}}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{5}{6}}=\dfrac{x+y-z}{\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{5}{6}}=\dfrac{-21}{\dfrac{7}{12}}=-36\\ \Rightarrow\left\{{}\begin{matrix}x=-36\cdot\dfrac{2}{3}=-24\\y=-36\cdot\dfrac{3}{4}=-27\\z=-36\cdot\dfrac{5}{6}=-30\end{matrix}\right.\)

1) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}=\frac{2x-4y+3z}{2.2-4.3+3.7}=\frac{-39}{13}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.2=-6\\y=-3.3=-9\\z=-3.7=-21\end{cases}}\)

2) \(9x=10y\Leftrightarrow\frac{x}{10}=\frac{y}{9},4y=3z\Leftrightarrow\frac{y}{9}=\frac{z}{12}\)

suy ra \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Leftrightarrow\hept{\begin{cases}x=6.10=60\\y=6.9=54\\z=6.12=72\end{cases}}\)

3) \(3x=4y=6z\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x-y+z}{4-3+2}=\frac{-9}{3}=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=-3.4=-12\\y=-3.3=-9\\z=-3.2=-6\end{cases}}\)

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

Bài 2:

a) \(\frac{4^2\cdot25^2+16\cdot125}{2^3\cdot5^2}=\frac{2^4\cdot5^4+2^4\cdot5^3}{2^3\cdot5^2}=\frac{2^4\cdot5^3\cdot\left(5+1\right)}{2^3\cdot5^2}=2\cdot5\cdot6=60\)

b) \(\frac{6^8\cdot2^4-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}=\frac{2^8\cdot3^8\cdot2^4-2^{10}\cdot2^4\cdot3^8}{3^9\cdot2^{12}-3^9\cdot2^{13}}=\frac{2^{12}\cdot3^8-2^{14}\cdot3^8}{3^9\cdot2^{12}\cdot\left(1-2\right)}=\frac{2^{12}\cdot3^8\cdot\left(1-2^2\right)}{3^9\cdot2^{12}\cdot\left(-1\right)}\)

\(=\frac{2^2-1}{3}=\frac{3}{3}=1\)

làm như điên mà chả chọn

mk nhầm sửa lại:

ta có:

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

\(5y=6z\)\(\Rightarrow\frac{y}{6}=\frac{z}{5}\)

\(\frac{x}{4}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\Rightarrow\)\(\frac{x}{24}=\frac{y}{18}=\frac{z}{15}\)

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

\(\frac{x^2}{24^2}=\frac{y^2}{18^2}=\frac{z^2}{15^2}=\frac{x^2+y^2+z^2}{24^2+18^2+15^2}=\frac{500}{1125}=\frac{4}{9}\)

\(\frac{x^2}{24^2}=\frac{4}{9}\Rightarrow x=\sqrt{\frac{4\cdot24^2}{9}}=16\)

\(\frac{y^2}{18^2}=\frac{4}{9}\Rightarrow y=\sqrt{\frac{4\cdot18^2}{9}}=12\)

\(\frac{z^2}{15^2}=\frac{4}{9}\Rightarrow z=\sqrt{\frac{15^2\cdot4}{9}}=10\)

Vậy x = 16, y = 12, z  = 10

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

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x+y-z}{2\cdot4+3-2}=\dfrac{9}{9}=1\)

Do đó: x=4; y=3; z=2