Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\)

=>x=15k; y=20k; z=24k

\(A=\dfrac{2\cdot15k+3\cdot20k+4\cdot24k}{3\cdot15k+4\cdot20k+2\cdot24k}=\dfrac{186}{173}\)

\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=\dfrac{2x+3y+4z}{30+60+96}=\dfrac{3x+4y+2z}{45+80+48}\\ \Leftrightarrow A=\dfrac{2x+3y+4z}{3x+4y+2z}=\dfrac{186}{173}\)

Ta có: \(\dfrac{x}{3}\)=\(\dfrac{y}{4}\) ; \(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)

=>\(\dfrac{x}{15}\)=\(\dfrac{y}{20}\)=\(\dfrac{z}{24}\)=k

=>x=15k

y=20k

z=24k

Thế x=15k; y=20k; z=24k vào biểu thức A, ta có:

\(\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)=\(\dfrac{30k+60k+96k}{45k+60k+120k}\)=\(\dfrac{k.\left(30+60+96\right)}{k.\left(45+60+120\right)}\)=\(\dfrac{186}{225}\)=\(\dfrac{62}{75}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

x/3=y/4 -> y=4x/3 (1)
y/5=z/6 -> y=5z/6 (2)

(1)+(2) -> x=5z/8 thay vào M=\(\dfrac{2.\dfrac{5z}{8}+3.\dfrac{5z}{6}+4z}{3.\dfrac{5z}{8}+4.\dfrac{5z}{6}+5z}\)=\(\dfrac{186}{245}\)

TÌM X Y Z

a) ta có : \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) = \(\dfrac{x}{16}=\dfrac{y}{24}\) ( 1)

\(\dfrac{y}{8}=\dfrac{z}{5}\) = \(\dfrac{y}{24}=\dfrac{z}{15}\) (2)

từ (1) và (2) , ta có : \(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}\)

mà x - y + z = 35

theo tính chất của dãy tỉ số bằng nhau , ta có :

\(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x-y+z}{16-24+15}=\dfrac{35}{7}=5\)

do đó : \(\dfrac{x}{16}=5\) => x = 5. 16 = 80

\(\dfrac{y}{24}=5\) => y = 5.24 = 120

\(\dfrac{z}{15}=5\) => z = 5.15 = 75

vậy x = 80

y = 120

z = 75

mấy câu còn lại thì tương tự nha bn

Câu b và câu e giống nhau

b)Từ 3x = 2y \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

Từ 5z = 2x \(\dfrac{z}{2}=\dfrac{x}{5}\) \(\Rightarrow\dfrac{x}{10}=\dfrac{z}{4}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{4}\)

Đặt \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=10k\\y=15k\\z=4k\end{matrix}\right.\)

\(\Rightarrow x\cdot y=10k\cdot15k=24\)

\(150\cdot k^2=24\)

\(k^2=\dfrac{4}{25}=\left(\pm\dfrac{2}{5}\right)^2\)

\(k=\pm\dfrac{2}{5}\)

Từ đó tìm được x ;y ;z bạn nhé

a/ Do \(x+y=22\Rightarrow y=22-x\)

\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)

\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)

\(\Leftrightarrow11x=88\Rightarrow x=8\)

\(\Rightarrow y=22-x=14\)

b/ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow y=\dfrac{4x}{3}\)

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

Vậy \(M=\dfrac{2x+3y+4z}{3x+4y+5z}=\dfrac{2x+3.\dfrac{4x}{3}+4.\dfrac{8x}{5}}{3x+4.\dfrac{4x}{3}+5.\dfrac{8x}{5}}\)

\(\Rightarrow M=\dfrac{x\left(2+4+\dfrac{32}{5}\right)}{x\left(3+\dfrac{16}{3}+8\right)}=\dfrac{\dfrac{62}{5}}{\dfrac{49}{3}}=\dfrac{186}{245}\)

Câu a:

Ta có: \(x+y=22\Rightarrow y=22-x\)

\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)

\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)

\(\Leftrightarrow11x=88\Rightarrow x=8\)

\(\Rightarrow y=22-x=22-8=14\)

Vậy \(x=8,y=14\)

\(\dfrac{4+x}{7+y}=\dfrac{4}{7}\Leftrightarrow7\left(4+x\right)=4\left(7+y\right)\Leftrightarrow7x=4y\Leftrightarrow x=\dfrac{4}{7}y.x+y=22\Rightarrow\dfrac{11}{4}x=22\Rightarrow x=8\Rightarrow y=14\)

Vậy.....

a)ta có 4+x/7+y=4/7

<=>7x+28=28+4y

<=> 7x=4y

lại có x+y=22

=>4/7y+y=22

<=>11/7y=22 <=> y=14

<=> x= 4/7*14=8

vậy x=8, y=14

b) Từ x/3=y/4 va y/5=z/6-->x/15=y/20=z/24 (1)
(1) = 2x/30=3y/60=4z/96=(2x+3y+4z)/186 (2) (t/c dãy tỉ số bằng nhau)
Ta lại có
(1) = 3x/45=4y/80=5z/120=(3x+4y+5z)/245 (3)(t/c dãy tỉ số bằng nhau)
Từ (2)(3) ta có(2x+3y+4z)/186=(3x+4y+5z)/245
Vậy M = (2x+3y+4z)/(3x+4y+5z)=186/245