Theo tính chất,ta có:

(x-1)*3=y*2.

=>(x-1)=2/3*y.

x+y=11.

=>(x-1)+y=10.

Từ đó làm tổng tỉ.

Theo tính chất ,ta có:

(x-1)x3=yx2

(x-1)=2/3xy

x+y=11

(x-1)+y=10

Từ đó làm tỉ số

1: xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

2: xy+x+6=0

=>x(y+1)=-6

=>(x;y+1)∈{(1;-6);(-6;1);(-1;6);(6;-1);(2;-3);(-3;2);(-2;3);(3;-2)}

=>(x;y)∈{(1;-7);(-6;0);(-1;5);(6;-2);(2;-4);(-3;1);(-2;2);(3;-3)}

3: -xy-x-y-1=0

=>xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

4: xy-x-y+1=0

=>x(y-1)-(y-1)=0

=>(x-1)(y-1)=0

=>\(\begin{cases}x-1=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=1\end{cases}\)

5: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>x(y+2)+(y+2)=-9

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}

6: ĐKXĐ: x<>0

\(\frac{5}{x}+\frac{y}{4}=\frac18\)

=>\(\frac{20+xy}{4x}=\frac18\)

=>\(\frac{40+2xy}{8x}=\frac{x}{8x}\)

=>40+2xy=x

=>x-2xy=40

=>x(1-2y)=40

=>x(2y-1)=-40

mà 2y-1 lẻ(do y nguyên)

nên (x;2y-1)∈{(-40;1);(40;-1);(8;-5);(-8;5)}

=>(x;2y)∈{(-40;2);(40;0);(8;-4);(-8;6)}

=>(x;y)∈{(-40;1);(40;0);(8;-2);(-8;3)}

8: (x+2)(y-3)=-3

=>(x+2;y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(-1;0);(-5;4);(-3;6);(1;2)}

(x - 1)/2 = y/3

=> 2y = 3x - 3

=> 2y + 3 = 3x

Ta có : x + y = 11

=> x = 11 - y

2y + 3 = 3x

=> 2y + 3 = 3 . (11 - y)

=> 2y + 3 = 33 - 3y

=> 5y = 30

=> y = 6

=> x = 11 - 6

=> x = 5

Vậy x = 5 và y = 6

xy+2x+y+11=0

=> x.(y+2)+y=-11

=> x.(y+2)+(y+2)= -11+2=-9

=> (x+1).(y+2)=-9

=> x+1 và y+2 thuộc Ư(-9)={1;-1;3;-3;9;-9}

x+1 y+2 x y 1 -9 0 -11 -1 9 -2 7 3 -3 2 -5 -3 3 -4 1 9 -1 8 -3 -9 1 -10 -1

Vậy....

\(xy+2x+y+11=0\)

\(\Rightarrow y\left(x+y\right)+2\left(x+5,5\right)=0\)

\(\Rightarrow\hept{\begin{cases}y\left(x+y\right)=0\\x+5,5=0\end{cases}\Rightarrow\hept{\begin{cases}y=0\\x=-5,5\end{cases}}}\)

A. dk \(\hept{\begin{cases}y+z+1\ne0\\x+z+1\ne0\\x+y\ne2\end{cases}}\)

Ap dung tinh chat day ti so bang nhau ta co

\(\frac{x}{y+z+1}=\frac{y}{x+z+1}\frac{z}{x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\) (1)

=> \(x+y+z=\frac{1}{2}\) (*) => y+z =1/2 - x

(1)  suy ra \(y+z+1=2x\)

<=> \(\frac{1}{2}-x+1=2x\Rightarrow x=\frac{1}{2}\)

thay vao (*) => y+z=0

tu (1) lai suy ra \(x+z+1=2y\)

<=> \(\hept{\begin{cases}z+y=0\\\frac{1}{2}+z+1=2y\end{cases}\Rightarrow\hept{\begin{cases}z=\frac{-1}{2}\\y=\frac{1}{2}\end{cases}}}\)

vay \(\left\{x;y;z\right\}=\left\{\frac{1}{2};\frac{1}{2};\frac{-1}{2}\right\}\)

b,     \(\left(x-11+y\right)^2+\left(x-y+4\right)^2=0\) 

<=> \(\hept{\begin{cases}x-11+y=0\\x-y-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\y=\frac{7}{2}\end{cases}}}\)

Vay \(\left\{x;y\right\}=\left\{\frac{15}{2};\frac{7}{2}\right\}\)

a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)

=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)

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

\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)

\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(48;12\right).\)

b)

\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)

⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)

⇒ \(\frac{5}{3}x=\frac{1}{21}\)

⇒ \(x=\frac{1}{21}:\frac{5}{3}\)

⇒ \(x=\frac{1}{35}\)

Vậy \(x=\frac{1}{35}.\)

\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

⇒ \(x-\frac{1}{2}=\frac{1}{3}\)

⇒ \(x=\frac{1}{3}+\frac{1}{2}\)

⇒ \(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}.\)

Có 1 câu bạn đăng mình làm ở dưới rồi mà.

Chúc bạn học tốt!

a)áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)

\(\)x/12=4 suy ra x=12.4=48

y/3=4 suy ra y=3.4 =12

b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)

\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)

\(\frac{5}{3}x=\frac{1}{21}\)

\(x=\frac{1}{21}:\frac{5}{3}\)

\(x=\frac{1}{35}\)

\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{1}{4}\)

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

\(x=\frac{-3}{20}\)

\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)

\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)

\(\left|x-\frac{2}{5}\right|=2\)

suy ra x-2/5=2 hoac x-2/5=-2

\(x-\frac{2}{5}=2\)

\(x=\frac{12}{5}\)

\(x-\frac{2}{5}=-2\)

\(x=\frac{-8}{5}\)

\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....

a) \(\frac{x}{y}=\frac{15}{7}\Leftrightarrow\)\(\frac{x}{15}=\frac{y}{17}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{y}{17}=\frac{x-2y}{15-2\cdot17}=\frac{16}{-19}\)

=> \(\begin{cases}x=-\frac{240}{19}\\y=-\frac{272}{19}\end{cases}\)

b) \(\frac{x}{y}=\frac{8}{11};\frac{z}{y}=\frac{3}{11}\)

\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{11};\frac{z}{3}=\frac{y}{11}\)

\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{11}=\frac{z}{3}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x}{8}=\frac{y}{11}=\frac{z}{3}=\frac{x+y-z}{8+11-3}=\frac{80}{16}=5\)

\(\Rightarrow\begin{cases}x=40\\y=55\end{cases}\)

c) \(\frac{x}{4}=\frac{y}{3}\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{11}\)

Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{11}=k\Rightarrow x=8k;y=6k;z=11k\)

Có \(xyz=-528\)

\(\Leftrightarrow8k\cdot6k\cdot11k=-528\)

\(\Leftrightarrow528\cdot k^3=-528\)

\(\Leftrightarrow k^3=-1\Leftrightarrow k=-1\)

Với k=-1 thì : x=-8;y=-6;x=-11

a) Từ \(\frac{x}{y}=\frac{15}{7}\Rightarrow\frac{x}{15}=\frac{y}{7}\)

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

\(\frac{x}{15}=\frac{y}{7}=\frac{x-2y}{15-14}=16\)

=> \(\begin{cases}x=240\\y=112\end{cases}\)

b) Từ \(\frac{x}{y}=\frac{8}{11}\Rightarrow\frac{x}{8}=\frac{y}{11}\)

\(\frac{z}{y}=\frac{3}{11}\Rightarrow\frac{z}{3}=\frac{y}{11}\)

=> \(\frac{x}{8}=\frac{y}{11}=\frac{z}{3}\)

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

\(\frac{x}{8}=\frac{y}{11}=\frac{z}{3}=\frac{x+y-z}{8+11-3}=\frac{80}{16}=5\)

=> \(\begin{cases}x=40\\y=55\\z=15\end{cases}\)

c)Từ \(\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{6}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{11}\)

Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{11}\) = k

=> \(\begin{cases}x=8k\\y=6k\\z=11k\end{cases}\)

=> x.y.z = -528 => 8k.6k.11k = -528 => 528k³ = -528

=> k³ = -1 => k = -1

=> \(\begin{cases}x=-8\\y=-6\\z=-11\end{cases}\)