Tiếc quá. Mik làm đc. Nhg mik chx = điện thọi nên k vt đc p/ số

ko sao

XL nha là x-y+z chứ ko phải là x+y+z

bài này ngày mai lên lớp Me chỉ cho

\(4x=3y=>\frac{x}{3}=\frac{y}{4}=>\frac{x}{15}=\frac{y}{20}\)

\(7y=5z=>\frac{y}{5}=\frac{z}{7}=>\frac{y}{20}=\frac{z}{28}\)

=>\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{x-y+z}{15-20+28}=\frac{-92}{23}=-4\)

=>x=-4.15=-60

=>y=-4.20=-80

=>z=-4.28=-112

Vậy x=-60,y=-80,z=-112

\(4x=2y\Rightarrow2x=y\Rightarrow\frac{x}{1}=\frac{y}{2};7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)

\(\frac{x}{5}=\frac{y}{10}=\frac{z}{14}=\frac{x-y+z}{5-10+14}=-\frac{46}{9}\)

\(x=5.\frac{-46}{9}=-\frac{230}{9}\)

\(y=10.\frac{-46}{9}=-\frac{460}{9}\)

\(z=14.\frac{-46}{9}=-\frac{644}{9}\)

a) thiếu đề bài 

4x-3y=7y-6x

<=> 4x+6y-3y-7y=0

<=>10x-10y=0

<=>x-y=0

Ta có x-y=0 và 2x+3y=55

2x+3y=55

<=>2x-2y+5y=55

<=>2(x-y)+5y=55

<=>5y=55 (x-y=0=>2(x-y)=0)

<=> y=11

x-y=0

=>x=y=11

Câu tiếp theo bạn tự làm nha!!

Chúc bạn học tốt!

cảm ơn bạn

4x=3y;7y=5z

⇒\(\frac{x}{4}\)=\(\frac{y}{3}\);\(\frac{y}{7}\)=\(\frac{z}{5}\)

⇒\(\frac{x}{28}=\frac{y}{21};\frac{y}{21}=\frac{z}{15}\)

⇒ \(\frac{x}{28}=\frac{y}{21}=\frac{z}{15}\)

⇒ \(\frac{x}{28}=\frac{y}{21}=\frac{z}{15}\) \(\frac{2x-3y+z}{56-63+15}=\frac{6}{8}=\frac{3}{4}\)

Tự tính tiếp

Ta có:

\(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\) (1)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}.\)

Có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}.\)

=> \(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\) và \(2x-3y+z=6.\)

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

\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x-3y+z}{30-60+28}=\frac{6}{-2}=-3.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{15}=-3\Rightarrow x=\left(-3\right).15=-45\\\frac{y}{20}=-3\Rightarrow y=\left(-3\right).20=-60\\\frac{z}{28}=-3\Rightarrow z=\left(-3\right).28=-84\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(-45;-60;-84\right).\)

Chúc bạn học tốt!

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

Ta có: \(4x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(7y=5z\)\(\Rightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\)\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)

Ta có: \(yz-2x^2=110\)

\(\Rightarrow20k.28k-2.\left(15k\right)^2=110\)

\(\Rightarrow560k^2-2.225k^2=110\)

\(\Rightarrow560k^2-450k^2=110\)

\(\Rightarrow k^2\left(560-450\right)=110\)

\(\Rightarrow110k^2=110\)

\(\Rightarrow k^2=1\)

\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

+) Khi k = 1, ta có: \(\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.1\\y=20.1\\z=28.1\end{cases}}\Rightarrow\hept{\begin{cases}x=15\\y=20\\z=28\end{cases}}\)

+) Khi k = -1, ta có: \(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.\left(-1\right)\\y=20.\left(-1\right)\\z=28.\left(-1\right)\end{cases}}\Rightarrow\hept{\begin{cases}x=-15\\y=-20\\z=-28\end{cases}}\)

Vậy...

Ta có: \(4x=3y\rightarrow\frac{x}{3}=\frac{y}{4}\rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)

          \(7y=5z\rightarrow\frac{y}{5}=\frac{z}{7}\rightarrow\frac{y}{20}=\frac{z}{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\left(k\varepsilonℕ^∗\right)\)

=> x  = 15k; y = 20k; z = 28k

Có: \(yz-2x^2=110\)

\(\Rightarrow20k\cdot28k-2\cdot(15k)^2=110\)

\(\Rightarrow560\cdot k^2-2\cdot225\cdot k^2=110\)

\(\Rightarrow560\cdot k^2-450\cdot k^2=110\)

\(\Rightarrow\left(560-450\right)\cdot k^2=110\)

\(\Rightarrow110\cdot k^2=110\)            \(\Rightarrow k^2=1\)

\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

\(x=15k\rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

\(y=20k\rightarrow\orbr{\begin{cases}y=20\\y=-20\end{cases}}\)

\(z=28k\rightarrow\orbr{\begin{cases}z=28\\z=-28\end{cases}}\)

Vậy...........................

thiếu chỗ \(\hept{\begin{cases}x=15k\\y=20k\end{cases}}\):)) thêm vào là \(\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)