Question

[Đào Khánh Huyền]

Bài mik để dưới phần trả lời ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a: \(\dfrac{2x^3+3x^2-2x-1}{2x+3}\)

\(=\dfrac{2x^3+3x^2-2x-3+2}{2x+3}\)

\(=\dfrac{x^2\left(2x+3\right)-\left(2x+3\right)+2}{2x+3}\)

\(=x^2-1+\dfrac{2}{2x+3}\)

vậy: \(2x^3+3x^2-2x-1=\left(x^2-1\right)\left(2x+3\right)+2\)

b: \(3x\left(x-2\right)+5\left(2-x\right)=0\)

=>3x(x-2)-5(x-2)=0

=>(x-2)(3x-5)=0

=>\(\left[{}\begin{matrix}x-2=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\)

Bài 2:

\(A=\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right)\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)

\(=\left(\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{2\left(x+y\right)}\right)\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)

\(=\dfrac{4xy+\left(x-y\right)^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}+\dfrac{y}{x-y}\)

\(=\dfrac{\left(x+y\right)^2\cdot2x}{2\left(x+y\right)^2\cdot\left(x-y\right)}+\dfrac{y}{x-y}\)

\(=\dfrac{x}{x-y}+\dfrac{y}{x-y}=\dfrac{x+y}{x-y}\)

b: \(\dfrac{x^2+y^2}{xy}=\dfrac{25}{12}\)

=>\(12\left(x^2+y^2\right)-25xy=0\)

=>\(12x^2-16xy-9xy+12y^2=0\)

=>\(4x\left(3x-4y\right)-3y\left(3x-4y\right)=0\)

=>\(\left(3x-4y\right)\left(4x-3y\right)=0\)

=>\(\left[{}\begin{matrix}3x-4y=0\\4x-3y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}y\\x=\dfrac{3}{4}y\end{matrix}\right.\)

x<y<0 nên \(x=\dfrac{4}{3}y\)

\(A=\dfrac{x+y}{x-y}=\dfrac{\dfrac{4}{3}y+y}{\dfrac{4}{3}y-y}=\dfrac{7}{3}:\dfrac{1}{3}=7\)

Answer 2

[Đào Khánh Huyền]