\(\dfrac{y}{0,4}\) chuyển thành y.\(\dfrac{5}{2}\)=\(\dfrac{y+z}{4}\)

Bài 11: Tìm x, y, z:

a) $x=4y=0,4\left(y+z\right)$ và $x+y+z=280$

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

1, 

trả lời 

a=b=1

cbht

Cho mk lời giải đầy đủ đi

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y-z=26\)

\(BCNN\left(3,5\right)=15\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}\)(1)

      \(\frac{y}{5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{15}=\frac{z}{12}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=\frac{26}{13}=2\)

\(\Rightarrow x=2.10=20\)

     \(y=2.15=30\)

     \(z=2.12=24\)

Vậy x = 20 ; y = 30 ; z = 24

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

b: \(B\ge2021\forall x,y\)

Dấu '=' xảy ra khi x=y=3