b. Áp dụng t/c dãy tỉ số = nhau:

\(\frac{x}{2}=\frac{y}{5}=\frac{x-y}{2-5}=-\frac{7}{3}\)

\(\Rightarrow\frac{x}{2}=-\frac{7}{3}\Leftrightarrow x=-\frac{7}{3}.2=-\frac{14}{3}\)

\(\Rightarrow\frac{y}{5}=-\frac{7}{3}\Leftrightarrow y=-\frac{7}{3}.5=-\frac{35}{3}\)

Vậy \(\hept{\begin{cases}x=-\frac{14}{3}\\y=-\frac{35}{3}\end{cases}}\)

c, Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k;y=3k;z=4k\)

Ta có: \(xyz=192\Leftrightarrow2k.3k.4k=192\)

                             \(\Leftrightarrow24k^3=192\)

                             \(\Leftrightarrow k^3=8\)

                             \(\Leftrightarrow k=2\)                          

\(\Rightarrow x=2.2=4\)  

    \(y=2.3=6\)

   \(z=2.4=8\)

e, Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{2x}{2}=\frac{3z}{9}\)

Áp dụng t/c dãy tỉ số = nhau:

\(\frac{2x}{2}=\frac{y}{2}=\frac{3z}{9}=\frac{2x-y+3z}{2-2+9}=\frac{10}{9}\)

\(\Rightarrow x=\frac{10}{9}\)

\(y=\frac{10}{9}.2=\frac{20}{9}\)

\(z=\frac{10}{9}.3=\frac{10}{3}\)

b,\(\frac{x}{2}=\frac{y}{5}=\frac{x-y}{2-5}=\frac{7}{-3}.\)

=>x= \(\frac{7}{-3}.2=-4\frac{2}{3}\)

y, \(\frac{7}{-3}.5=-11\frac{2}{3}\)

khó thế

đáp án A 100^o

k cho minh nha

chuc ban hoc tot

Phải có lời giải

- Vì \(\frac{x}{5}=\frac{y}{3}\)=) \(3x=5y\)=) \(x=\frac{5y}{3}\)
=) \(x^2-y^2=4\)=) \(\left(\frac{5y}{3}\right)^2-y^2=4\)
=) \(\frac{25y^2}{9}-y^2=4\)=) \(\frac{25y^2}{9}-\frac{9y^2}{9}=\frac{36}{9}\)
=) \(25y^2-9y^2=36\)=) \(16y^2=36\)=) \(y^2=\frac{36}{16}=\frac{9}{4}\frac{3^2}{2^2}\)=) \(y=\frac{3}{2}\)
=) \(x=\frac{5.\frac{3}{2}}{3}=\frac{\frac{15}{2}}{3}=\frac{5}{2}\)

a) Đặt x/5 = y/3 = k => x = 5k ; y = 3k

Ta có: x² - y² = 4

=> (5k)² - (3k)² = 4

=> 25k² - 9k² = 4

=> 16k² = 4

=> k² = 1/4

=> k = ±1/2

Với k = 1/2 thì x = 5/2, y = 3/2

Với k = -1/2 thì x = -5/2, y = -3/2

b) Theo tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

=> x + y + z = 1/2 ; x/y+z+1 = 1/2 ; y/z+x+1 = 1/2 ; z/x+y-2 = 1/2

=> \(\hept{\begin{cases}y+z+1=2x\\z+x+1=2y\\x+y-2=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+1=3y\\\frac{1}{2}-2=3z\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{cases}}\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)

\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)

\(\frac{2}{3}-x=-\frac{7}{6}\)

\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)

\(x=\frac{2}{3}+\frac{7}{6}\)

\(x=\frac{11}{6}\)

A B C D E F

A B C D E