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​=

chết rùi nó bị lỗi

xin lỗi nha

a, x=5y/7thay vào x-y = 1

 ta đc:

 5y/7 - y=1

<=> -2y=7

<=> y=-7/2

=> x=-5/2

3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)

4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)

5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)

6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)

7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)

Câu 3:

\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)

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

\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)

Câu b:

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)

Câu c:

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

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)

Câu d:

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

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)

Câu e:

\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)

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

\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)

\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)

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

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

a) ADTCDTSBN

có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)

=> x/2 = 3 => x = 6

y/3 = 3 => y = 9

z/4 = 3 => z = 12

KL:...

b,c làm tương tự nha

d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)

ADTCDTSBN

có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)

=>...

e) ADTCDTSBN

có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)

\(=\frac{21+6}{9}=\frac{27}{9}=3\)

=>...

g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)

mà xy = 12 => 4k.3k = 12

                          12.k² = 12

                              k² = 1

                        => k = 1 hoặc k = -1

=> x = 4.1 = 4

y = 3.1 = 3

x=4.(-1) = -4 

y=3.(-1) = -3

KL:...

h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)

=>...

b) x/3=y/4; y/5=z/7 va 2x+ 3y- z= 186 
Ta có 
x/3=y/4 <=> x/15=y/20 (1) 
y/5=z/7 <=> y/20=z/28 (2) 
Từ (1) và (2) suy ra 
x/15=y/20=z/28 
<=> 2x/30=3y/60=z/28 = (2x+3y-z)/(30+60-28) = 186/62 = 3 
Vậy: 
x=3.15=45 
y=3.20=60 
z=3.28=84 

Ko bít làm!