1a) \(0,31:0,91=x:\frac{49}{3}\)

=> \(\frac{0,31}{0,91}=\frac{3x}{49}\)

=> \(3x=\frac{3}{7}.49\)

=> \(3x=21\)

=> \(x=21:3=7\)

b) \(6,88:x=12:27\)

=> \(\frac{6,88}{x}=\frac{12}{27}\)

=> \(x=6,88:\frac{4}{9}\)

=> \(x=15,48\)

c) \(\frac{25}{3}:\frac{35}{3}=13:2x\)

=> \(\frac{13}{2x}=\frac{5}{7}\)

=> \(2x=13:\frac{5}{7}\)

=> \(2x=\frac{91}{5}\)

=> \(x=\frac{91}{5}:2=\frac{91}{10}\)

d) \(\left(x-1\right):24,5=5:8,75\)

=> \(\frac{x-1}{24,5}=\frac{5}{8,75}\)

=> \(x-1=\frac{4}{7}.24,5\)

=> \(x-1=14\)

=> \(x=14+1=15\)

2a) Ta có: \(\frac{x}{y}=\frac{5}{7}\) => \(\frac{x}{5}=\frac{y}{7}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

   \(\frac{x}{5}=\frac{y}{7}=\frac{x+y}{5+7}=\frac{4,08}{12}=0,34\)

=> \(\hept{\begin{cases}\frac{x}{5}=0,34\\\frac{y}{7}=0,34\end{cases}}\) => \(\hept{\begin{cases}x=0,34.5=1,7\\y=0,34.7=2,38\end{cases}}\)

Vậy x = 1,7; y = 2,38

b) Ta có: \(\frac{x}{y}=-\frac{3}{7}\) => \(\frac{x}{-3}=\frac{y}{7}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

  \(\frac{x}{-3}=\frac{y}{7}=\frac{x-y}{-3-7}=\frac{-40}{-10}=4\)

=> \(\hept{\begin{cases}\frac{x}{-3}=4\\\frac{y}{7}=4\end{cases}}\) => \(\hept{\begin{cases}x=4.\left(-3\right)=-12\\y=4.7=28\end{cases}}\)

vậy x = -12; y = 28

c) Ta có: \(\frac{x}{y}=\frac{3}{5}\) => \(\frac{x}{3}=\frac{y}{5}\)

Đặt : \(\frac{x}{3}=\frac{y}{5}=k\) => \(\hept{\begin{cases}x=3k\\y=5k\end{cases}}\) (*)

Khi đó, ta có: xy = 1215

hay 3k. 5k = 1215

=> 15k² = 1215

=> k² = 1215 : 15 = 81

=> k = \(\pm\)9 

Thay k = \(\pm\)9 vào (*), ta được:

+) x = 3. (\(\pm\)9) = \(\pm\)27

+) y = 5. (\(\pm\)9) = \(\pm\)45

Vậy ...

Mơn bn nhiều <3

Nếu \(x\)là \(9\)phần thì \(y\)là \(14\)phần.

Tổng số phần bằng nhau là: 

\(9+14=23\)(phần) 

\(x\)bằng: 

\(69\div23\times9=27\)

\(y\)bằng: 

\(69-27=42\)

a: k=3/2

b: k=6

Đặt A=\(\left(x-5\right)\left(y+40\right)\)

\(a,x=5;y=1000\)

\(A=\left(5-5\right)\left(1000+40\right)\)

\(A=0.1040\)

\(A=0\)

\(b,x=-10;y=15\)

\(A=\left(-10-5\right)\left(15+40\right)\)

\(A=\left(-15\right).55\)

\(A=-825\)

\(c,x=16;y=-50\)

\(A=\left(16-5\right)\left(-50+40\right)\)

\(A=11.\left(-10\right)\)

\(A=-110\)

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