Tìm số hữu tỉ x, y, z biết x(x+y+z) = -5; y(x + y + z)= 9; z(x + y + z)= 5

Bài 1:

\(A=\dfrac{\left(1,09-0,29\right)\cdot\dfrac{5}{4}}{\left(18,9-16,65\right)\cdot\dfrac{8}{9}}=\dfrac{0,8\cdot1,25}{2,25\cdot\dfrac{8}{9}}=\dfrac{1}{2}\)

\(B=\left[0,8\cdot7+\left(0,8\right)^2\right]\left(1,25\cdot7-\dfrac{4}{5}\cdot1,25\right)+31,64\)

\(=0,8\cdot\left(7+0,8\right)\cdot1,25\left(7-0,8\right)+31,64\)

\(=0,8\cdot7,8\cdot1,25\cdot6,2+31,64\)

\(=6,24\cdot7,75+31,64=48,36+31,65=80\)

\(\Rightarrow A:B=\dfrac{1}{2}:80=\dfrac{1}{160}\)

Vậy A gấp 1/160 lần B

bài 2:

\(\dfrac{x}{4}-\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{4}-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{x-2}{4}\)

=>y(x-2)=4

=>y và x-2 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Ta có bẳng:

Vậy....

bài 3:

Ta có: x-y=x:y => x=xy+y=y(x+1) => x:y=y(x+1):y=x+1 (1)

Mà x:y=x-y (2)

Từ (1) và (2) => y = -1

Lại có: x=y(x+1) => x=(-1)(x+1) => x=-x-1 => 2x=-1 => x=\(\dfrac{-1}{2}\)

Vậy x=-1/2, y=-1

bài 4:

Ta có: x(x+y+z)+y(x+y+z)+z(x+y+z)=-5+9+5

=>(x+y+z)²=9

=>x+y+z=3 hoặc x+y+z=-3

Nếu x+y+z=3 => \(\left\{{}\begin{matrix}3x=-5\\3y=9\\3z=5\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=3\\z=\dfrac{5}{3}\end{matrix}\right.\)

Nếu x+y+z=-3 => \(\left\{{}\begin{matrix}-3x=-5\\-3y=9\\-3z=5\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-3\\z=\dfrac{-5}{3}\end{matrix}\right.\)

Vậy....