`y=2/3x`

`=>3y=2x`

`=>8x=12y`

Mặt khác:`4z=3y`

`=>z=3/4y`

`=>5z=15/4y`

Thay `8x=12y,5z=15/4y` vào `8x+9y+5z=1980`

`=>15/4y+9y+12y=1980`

`=>21y+15/4y=1980`

`=>99/4y=1980`

`=>1/4y=20`

`=>y=80`

`=>x=3/2y=120,z=3/4y=60`

Vậy `(x,y,z)=(120,80,60)`

Ta có: 4z=3y

nên \(4z=3\cdot\dfrac{2}{3}x=x\)

hay \(z=\dfrac{1}{4}x\)

Ta có: 8x+9y+5z=1980

\(\Leftrightarrow8x+9\cdot\dfrac{2}{3}x+5\cdot\dfrac{1}{4}x=1980\)

\(\Leftrightarrow x\cdot\dfrac{61}{4}=1980\)

hay \(x=\dfrac{7920}{61}\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}x=\dfrac{2}{3}\cdot\dfrac{7920}{61}=\dfrac{5280}{61}\\4z=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\4z=\dfrac{15840}{61}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\z=\dfrac{3960}{61}\end{matrix}\right.\)

Vậy: \(\left(x,y,z\right)=\left(\dfrac{7920}{61};\dfrac{5280}{61};\dfrac{3960}{61}\right)\)

Giải:

Ta có:

\(y=\dfrac{2}{3}x;4z=3y\) 

\(\Rightarrow x=\dfrac{3}{2}y;z=\dfrac{3}{4}y\)

Theo đề bài ta có:

\(8x+9y+5z=1980\) 

\(\Rightarrow8.\dfrac{3}{2}y+9y+5.\dfrac{3}{4}y=1980\) 

\(\Rightarrow12y+9y+\dfrac{15}{4}y=1980\)  

\(\Rightarrow y.\left(12+9+\dfrac{15}{4}\right)=1980\) 

\(\Rightarrow y.\dfrac{99}{4}=1980\) 

\(\Rightarrow y=1980:\dfrac{99}{4}\) 

\(\Rightarrow y=80\) 

\(\Rightarrow x=\dfrac{3}{2}.80=120\) 

\(\Rightarrow z=\dfrac{3}{4}.80=60\) 

Vậy \(\left(x;y;z\right)=\left(120;80;60\right)\) 

chiu roi

ban oi

tk nhe@@@@@@@@@@

ai tk minh minh tk lai!!

mik k cho bạn 1 cái

minh lam cau b) roi dc co 2/3 thoy ban tham khao nhe phan () la minh giai thich nha dung viet vo bai !!
2x=3y ; 5y = 7z

+) 10x=15y=21z   ( Quy dong)

+)10x/210 = 15y/210 = 21z/210       ( BC)

+) x/21 = y/14 = z/10  ( Rut gon)

+) 3x/63 = 7y/98 =  5z/50 = 3x-7y+ 5z / 63 - 98 - 50 = -30/14 = -2

+ x/21 = 2 => ............  phan nay minh chua xong neu xong thi minh pm not cho

2x=3y,5y=7z và 3x-7y+5z=30

cho mình

b) Ta có: \(\text{10x=6y=5z}\Rightarrow\frac{10x}{30}=\frac{6y}{30}=\frac{5z}{30}\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{6}\) và \(x+y-z=24\)

Áp dụng t/c dãy tỉ số = nhau, ta có: 

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{6}=\frac{x+y-z}{3+5-6}=\frac{24}{2}=12\)

Khi đó:  \(\frac{x}{3}=12\Rightarrow x=36\)

            \(\frac{y}{5}=12\Rightarrow y=60\)

             \(\frac{z}{6}=12\Rightarrow z=72\)

Vậy\(x=36\)      :\(y=60\)          \(z=72\)  

thank you

Với các bài khá nâng cao như vậy bạn đăng tách ra nhé!

Answer:

a) Ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Ta đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

Ta có: \(5z^2-3x^2-2y^2=594\)

\(\Rightarrow5.\left(5k\right)^2-3.\left(3k\right)^2-2.\left(4k\right)^2=594\)

\(\Rightarrow5.5^2k^2-3.3^2k^2-2.4^2k^2=594\)

\(\Rightarrow5.25k^2-3.9k^2-2.16.k^2=594\)

\(\Rightarrow125k^2-27k^2-32k^2=594\)

\(\Rightarrow k^2.\left(125-27-32\right)=594\)

\(\Rightarrow k^2.66=594\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

Với \(k=3\Rightarrow\hept{\begin{cases}x=3.3=9\\y=3.4=12\\z=3.5=15\end{cases}}\)

Với \(k=-3\Rightarrow\hept{\begin{cases}x=\left(-3\right).3=-9\\y=\left(-4\right).3=-12\\z=\left(-5\right).3=-15\end{cases}}\)

Answer:

b) \(3.\left(x-1\right)=2.\left(y-2\right)\Rightarrow6.\left(x-1\right)=4.\left(y-2\right)\)

Mà: \(4.\left(y-2\right)=3.\left(z-3\right)\)

\(\Rightarrow6.\left(x-1\right)=4.\left(y-2\right)=3.\left(z-3\right)\)

\(\Rightarrow\frac{6.\left(x-1\right)}{12}=\frac{4.\left(y-2\right)}{12}=\frac{3.\left(z-3\right)}{12}\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}==\frac{\left(2x-2\right)+\left(3y-6\right)-z}{4+9-4}=\frac{2x-2+3y-6-z}{9}=\frac{\left(2x+3y-z\right)-\left(2+6\right)}{9}=\frac{50-8}{9}=\frac{14}{3}\)

\(\Rightarrow\hept{\begin{cases}x-1=2.\frac{14}{3}=\frac{28}{3}\\y-2=3.\frac{14}{3}=14\\z-3=4.\frac{14}{3}=\frac{56}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{31}{3}\\y=16\\z=\frac{68}{3}\end{cases}}\)

c) \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y-z}{18+16-15}=\frac{38}{19}=2\)

\(\Rightarrow\frac{x}{18}=2\Rightarrow x=18.2=36\)

\(\Rightarrow\frac{y}{16}=2\Rightarrow y=16.2=32\)

\(\Rightarrow\frac{z}{15}=2\Rightarrow z=15.2=30\)

\(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)

\(\Leftrightarrow\dfrac{4\left(5z-6y\right)}{16}=\dfrac{5\left(6x-4z\right)}{25}=\dfrac{6\left(4y-5x\right)}{36}\)

\(\Leftrightarrow\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}\)

ADTCDTSBN có:

\(\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}=\dfrac{20z-24y+30x-20z+24y-30x}{16+25+36}=0\)

Do đó \(20z-24y=0;30x-20z=0\)

\(\Leftrightarrow5z=6y;6x=4z\)

\(\Rightarrow y=\dfrac{5z}{6};x=\dfrac{4z}{6}\)

Có \(3x-3y+5z=96\Rightarrow3.\dfrac{4z}{6}-3.\dfrac{5z}{6}+5z=96\)

\(\Rightarrow z=\dfrac{64}{3}\) \(\Rightarrow y=\dfrac{160}{9}\)và \(x=\dfrac{128}{9}\)

Vậy...

cho x/3 = y/4 và y/5 = z/6. tìm M = 2x + 3y+ 4z / 3x + 4y + 5z

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)

nên \(\dfrac{x}{15}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{6}\)

nên \(\dfrac{y}{20}=\dfrac{z}{24}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)(3)

\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{4z}{96}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{4z}{96}=\dfrac{2x+3y+4z}{30+60+96}=\dfrac{2x+3y+4z}{186}\)

Từ (3) suy ra \(\dfrac{3x}{45}=\dfrac{4y}{80}=\dfrac{5z}{120}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x}{45}=\dfrac{4y}{80}=\dfrac{5z}{120}=\dfrac{3x+4y+5z}{45+80+120}=\dfrac{3x+4y+5z}{245}\)

Suy ra: \(M=\dfrac{2x+3y+4z}{3x+4y+5z}=\dfrac{186}{245}\)