ta có: \(\frac{4x}{5}=\frac{5y}{6}\Rightarrow24x=25y\Rightarrow\frac{x}{25}=\frac{y}{24}\Rightarrow\frac{x}{50}=\frac{y}{48}\)

\(\frac{3y}{8}=\frac{2z}{7}\Rightarrow21y=16z\Rightarrow\frac{y}{16}=\frac{z}{21}\Rightarrow\frac{y}{48}=\frac{z}{63}\)

\(\Rightarrow\frac{x}{50}=\frac{y}{48}=\frac{z}{63}=\frac{10x}{500}=\frac{4y}{192}=\frac{3z}{189}\)

ADTCDTSBN

có:\(\frac{3z}{189}=\frac{4y}{192}=\frac{10x}{500}=\frac{3z+4y-10x}{189+192-500}=\frac{-238}{-119}=2\)

=> x/50 = 2 => x = 100

y/48 = 2 => y = 96

z/63 = 2 => z = 126

KL:...

bạn ơi đề thiếu

x = -19 x 3 : (-57) = 1

y = -19 x (-9) : (-57) = 3

z = -57 x (-8) : (-19) = 24

Tk mình nha mọi người!!!!

Ta có:

    \(\frac{x}{3}=\frac{y}{-9}=\frac{8}{-z}=\frac{-19}{-57}\) 

\(\Rightarrow\frac{x}{3}=\frac{-y}{9}=\frac{-8}{z}=\frac{19}{57}\)

\(\Rightarrow\frac{x}{3}=\frac{19}{57}\)

\(\Rightarrow x\times57=19\times3=57\)

\(\Rightarrow x=57\div57=1\)

\(\Rightarrow\frac{1}{3}=\frac{-y}{9}\)

\(\Rightarrow-y\times3=1\times9=9\)

\(\Rightarrow-y=9\div3=3\)

\(\Rightarrow y=-3\)

\(\Rightarrow\frac{-3}{-9}=\frac{3}{9}=\frac{-8}{z}\)

\(\Rightarrow3\times z=-8\times9=-72\)

\(\Rightarrow z=-72\div3=-24\)

Vậy các giá trị x,y,z thỏa mãn đề bài là:

x=1;y=-3,z=-24

x=1,y=-3,z=-24 vì -19/-57=1/3 

\(\frac{x}{8}-\frac{1}{y}=\frac{1}{4}\)

=\(\frac{x:2}{8:2}-\frac{1}{y}=\frac{1}{4}\)

=\(\frac{x:2}{4}-\frac{1}{y}=\frac{1}{4}\)

để cho cùng mẫu thì y=4

vậy ta có

\(\frac{x:2}{4}-\frac{1}{4}=\frac{1}{4}\)

chuyển vế \(\frac{1}{4}\)đổi dấu rồi tính ta có

\(\frac{x:2}{4}=\frac{2}{4}\)

vậy x=\(2\times2=4\)

đs: y=4;x=4

\(2y=3z\)

\(=>\frac{y}{3}=\frac{z}{2}\)

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

\(=\frac{x+y+z}{2+3+2}\)(tính chất dãy tỉ số bằng nhau)

\(=\frac{49}{7}\)

\(=7\)

\(=>x=7.2=14,y=7.3=21,z=7.2=14\)

a) ĐKXĐ: \(x\ne-1\)

Ta có:

\(\frac{x+1}{8}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=8^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}\left(TMĐKXĐ\right)}\)

\(\)

a, \(\frac{x+1}{8}=\frac{8}{x+1}\)

\(\Leftrightarrow\left(x+1\right)^2=8.8\)

\(\Leftrightarrow\left(x+1\right)=\pm8\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}}}\)

b, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\left(2x+3y=186\right)\)

Theo đề bài ta có:

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3.5}=\frac{y}{4.5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5.4}=\frac{z}{7.4}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

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

\(\frac{x}{15}=\frac{y}{20}=\frac{2x}{30}=\frac{3y}{60}=\frac{2x+3y}{90}=\frac{186}{90}=\frac{31}{15}\)

\(\Rightarrow\frac{2x}{30}=\frac{31}{15}\Rightarrow2x=62\Rightarrow x=31\)

\(\frac{3y}{60}=\frac{31}{15}\Rightarrow3y=124\Rightarrow y=\frac{124}{3}\)

Mà \(\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{\frac{124}{3}}{20}=\frac{z}{28}\Rightarrow\frac{31}{15}=\frac{z}{28}\)

Từ đây bạn tìm nốt z nha

\(\text{Đặt }\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\\z=7k\end{cases}}\)

\(\left(x-z\right)^3=\left(3k-7k\right)^3=\left(-4k\right)^3\)

\(8.\left(x-y\right)^2.\left(y-z\right)=8.\left(3k-5k\right)^2.\left(5k-7k\right)=32k^2.\left(-2\right)k=-4k^3\)

=> đpcm