`y+7/2xxy+yxx3/5=102`

`y+7/2xxyxx1xx+yxx3/5=102`

`yxx(7/2+1+3/5)=102`

`yxx5,1=102`

`y=102:5,1`

`y=20`

Vậy..

`@An`

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:

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

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

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

câu cuối làm tương tự

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)
2) \(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

3) \(\dfrac{x}{-3}=\dfrac{4}{y}\Rightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)

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

\(\frac{y}{5}=\frac{z}{7}\\ \Rightarrow\frac{y}{10}=\frac{z}{14}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{14}=\frac{2x}{20}=\frac{3y}{45}=\frac{z}{14}=\frac{2x+3y+z}{20+45+14}=\frac{102}{79}\)

\(\Rightarrow x=\frac{1020}{79};y=\frac{1530}{79};z=\frac{1428}{79}\)

suy ra:  x/10 = y/15  ;   y/15 = z/21 và 2x +3y +z =102

suy ra: x/10 = y/15 = z/21 và  2x +3y +z =102

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

x/10=y/15=z/21 = 2x/20 = 3y/45 = z/21 = 2x+3y+z / 20 +45 +21 = 102/86 = 51/43 

SAI ĐỀ RÙI

a/ 2x = 5y và x - 2y = -12

Ta có: 2x = 5y => \(\frac{x}{5}=\frac{y}{2}\)

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

\(\frac{x}{5}=\frac{y}{2}=\frac{x-y}{5+2}=\frac{x-2y}{5+2.2}=\frac{-12}{9}=-\frac{4}{3}\)

\(\frac{x}{5}=-\frac{4}{3}\Rightarrow x=\frac{-4}{3}.5=-\frac{20}{3}\)

\(\frac{y}{2}=-\frac{4}{3}\Rightarrow y=-\frac{4}{3}.2=-\frac{8}{3}\)

Vậy:.................

b/ 2x = 3y = 4z và x + y + z =21

Ta có: 2x = 3y = 4z

=> \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\)

=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

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

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)

\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{21}{13}.6=\frac{126}{13}\)

\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{21}{13}.4=\frac{84}{13}\)

\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{21}{13}.3=\frac{63}{13}\)

Vậy:...............

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

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{32}{8}=4\)

\(\frac{x}{3}=4\Rightarrow x=4.3=12\)

\(\frac{y}{5}=4\Rightarrow y=4.5=20\)

Vậy:................

d/ Ta có: 7x = 3y

=> \(\frac{7x}{21}=\frac{3y}{21}\)

=> \(\frac{x}{3}=\frac{y}{7}\)

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

\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)

\(\frac{x}{4}=-4\Rightarrow x=\left(-4\right).4=-16\)

\(\frac{y}{7}=-4\Rightarrow y=\left(-4\right).7=-28\)

Vậy:................

1,\(2x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{2}\Rightarrow\frac{x}{5}=\frac{2y}{4}\)

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

\(\frac{x}{5}=\frac{2y}{4}=\frac{x-2y}{5-4}=\frac{-12}{1}=-12\)

Do đó:

\(\frac{x}{5}=-12\Rightarrow x=-60\)

\(\frac{2y}{4}=-12\Leftrightarrow\frac{y}{2}=-12\Rightarrow x=-24\)

Vây x = -60,y = -24

2, 2x = 3y = 4z \(\Rightarrow BCNN\left(2;3;4\right)=12\)

nên \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

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

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)

Do đó

\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{6.21}{13}=\frac{126}{13}\)

\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{4.21}{13}=\frac{84}{13}\)

\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{3.21}{13}=\frac{63}{13}\)

f/ \(\frac{x}{3}=\frac{y}{5};\frac{y}{2}=\frac{z}{7}\)

=> \(\frac{x}{6}=\frac{y}{10};\frac{y}{10}=\frac{z}{35}\)

=> \(\frac{x}{6}=\frac{y}{10}=\frac{z}{35}\)

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

\(\frac{x}{6}=\frac{y}{10}=\frac{z}{35}=\frac{x+y+z}{6+10+35}=\frac{102}{51}=2\)

\(\frac{x}{6}=2\Rightarrow x=2.6=12\)

\(\frac{y}{10}=2\Rightarrow y=2.10=20\)

\(\frac{z}{35}=2\Rightarrow z=2.35=70\)

Vậy:.................

h/ Đăt: \(\frac{x}{3}=\frac{y}{16}=k\)

\(\frac{x}{3}=k\Rightarrow x=3k\)

\(\frac{y}{16}=k\Rightarrow y=16k\)

Ta có: x. y = 192

=> 3k. 16k = 192

=> k². (3. 16) = 192

=> k². 48 = 192

=> k² = 192 : 48 = 4

=> k = \(\pm\) 2

*Với k = 2

\(\frac{x}{3}=k\Rightarrow x=3.k=3.2=6\)

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

*Với k = -2

\(\frac{x}{3}=k\Rightarrow x=3.k=3.\left(-2\right)=-6\)

\(\frac{y}{16}=k\Rightarrow y=16.k=16.\left(-2\right)=-32\)

Vậy:..........

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ