Bài 1: 

Ta có:

\(y-x=25\Rightarrow y=25+x\)

Mà \(7x=4y\Rightarrow7x=4\cdot\left(25+x\right)\)

\(7x=100+4x\)

\(\Rightarrow7x-4x=100\)

\(3x=100\)

\(x=\frac{100}{3}\)

bài 1 :

Ta có: 7x=4y ⇔ x/4=y/7

áp dụng tính chất dãy tỉ số bằng nhau ta có 

x/4=y/7=(y-x)/(7-4)=100/3

⇒x= 4 x 100/3=400/3 ; y = 7 x 100/3=700/3

bài 2 

ta có x/5 = y/6 ⇔ x/20=y/24

         y/8 = z/7 ⇔ y/24=z/21

⇒x/20=y/24=z/21

ADTCDTSBN(bài 1 có)

x/20=y/24=z/21=(x+y)/(20+24)=69/48=23/16

⇒x= 20 x 23/16 = 115/4

   y= 24x 23/16=138/2

   z=21x23/16=483/16

bài 3

x:y:z=3:8:5  ⇔ x/3=y/8=z/5

ADTCDTSBN

x/3=y/8=z/5=(3x+y-4z)/(9+8-10)=14/7=2

⇒x = 2x3 = 6 ; y= 2x8=`16; z=2x5=10

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=0,97,15​z=1,11,4​.x=1,11,4​.0,96,05​=0,998,47​​

d) �2=�3=�5;���=−302x​=3x​=5z​;xyz=−30

�2=�3=�5=���2.3.5=−3030=−12x​=3x​=5z​=2.3.5xyz​=30−30​=−1

⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧​x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5​
 

chết rùi nó bị lỗi

xin lỗi nha

gọi x-1/2=y+3/4=z-5/6=k =>x=k+1/2; y=k-3/4; z=k+5/6

ta co: 5z-3x-4y=[5(k+5/6)]-[3(k+1/2)]-4[(k-3/4)]=[5k+25/6]-[3k+3/2]-[4k-3]=5k+25/6-3k-3/2-4k+3=(5k-3k-4k)+(25/6+3-3/2)=-2k+34/9=50

roi con la tu lam nha bye

no biết

ko bt àk

ai bt ko tl hộ mk ik

Ta có:

\(3x=4y\Leftrightarrow\frac{x}{4}=\frac{y}{3}\)  và \(y-x=5\)

Áp dụng tính chất của dạy tỉ số bằng nhau:

\(\frac{x}{4}=\frac{y}{5}=\frac{y-x}{5-4}=\frac{5}{1}=5\)

\(\hept{\begin{cases}\frac{x}{4}=5\Rightarrow x=5.4=20\\\frac{y}{5}=5\Rightarrow y=5.5=25\end{cases}}\)

Vậy \(x=20;y=25\)

b)

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}\) và \(a-2b+3c=35\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a-2b+3c}{3-2.4+3.5}=\frac{35}{10}=3,5\)

\(\hept{\begin{cases}\frac{a}{3}=3,5\Rightarrow a=3,5.3=10,5\\\frac{b}{4}=3,5\Rightarrow b=3,5.4=14\\\frac{c}{5}=3,5\Rightarrow c=3,5.5=17,5\end{cases}}\)

Vậy   \(a=10,5;b=14;c=17,5\)

Bài 1: \(3x=4y\Leftrightarrow y=\frac{3x}{4}\)

thay vào \(y-x=5\Leftrightarrow\frac{3x}{4}-x=5\Leftrightarrow\frac{-x}{4}=5\Leftrightarrow x=-20\Leftrightarrow y=\frac{3x}{4}=\frac{3.\left(-20\right)}{4}\)=-15

Bài 2: Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{2b}{8}=\frac{3c}{15}=\frac{a-2b+3c}{3-8+15}=\frac{35}{10}=\frac{7}{2}\)

=>\(a=\frac{7}{2}.3=\frac{21}{2};b=\frac{7}{2}.4=14;c=\frac{7}{2}.5=\frac{35}{2}\)

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}$

$=\frac{5z-25-3x+3-4y+12}{30-6-16}$

$=\frac{\left(5z-3x-4y\right)-\left(25-3-12\right)}{8}$

$=\frac{50-10}{8}=\frac{40}{8}=5$

+) $\frac{x-1}{2}=5\Rightarrow x=11$

+) $\frac{y+3}{4}=5\Rightarrow y=17$

+) $\frac{z-5}{6}=5\Rightarrow z=35$

Vậy bộ số $\left(x;y;z\right)$ l

à $\left(11;17;35\right)$

Ta có : \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\Rightarrow\frac{-3x+3}{-6}=\frac{-4y-12}{-16}=\frac{5z-25}{30}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{-3x+3}{-6}=\frac{-4y-12}{-16}=\frac{5z-25}{30}=\frac{-3x-4y+5z+3-12-25}{8}=2\)

\(\Rightarrow-3x+3=-12\Leftrightarrow-3x=-15\Leftrightarrow x=5\)

\(\Rightarrow-4y-12=-32\Leftrightarrow-4y=-20\Leftrightarrow y=5\)

\(\Rightarrow5z-25=60\Leftrightarrow z=17\)

sửa hộ mình y = - 5 bạn nhé