a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)

\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)

c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)

\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)

d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)

f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

ĐKXĐ: \(2x-y-1\ge0;x+2y\ge0\)

Đặt \(\sqrt{2x-y-1}=a;\sqrt{x+2y}=b\left(a,b\ge0\right)\). Khi đó ta có:

\(\left(2b^2-1\right)a=\left(2a^2-1\right)b\Leftrightarrow\left(a-b\right)\left(2ab+1\right)=0\)

\(\Leftrightarrow a=b\) hoặc \(2ab+1=0\)(loại vì \(a,b\ge0\))

Suy ra: \(\sqrt{2x-y-1}=\sqrt{x+2y}\Leftrightarrow x=3y+1\)

Pt đầu tiên trở thành: \(\left(3y+1\right)^2-5y^2-8y=3\)

\(\Leftrightarrow\left(y-1\right)\left(2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}y=1\\y=-\frac{1}{2}\end{cases}}\)

+) Với  \(y=1\Rightarrow x=4\Rightarrow\left(x;y\right)=\left(4;1\right)\)(tm)

+) Với  \(y=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}\Rightarrow\left(x;y\right)=\left(-\frac{1}{2};-\frac{1}{2}\right)\) (loại)

Vậy hpt có nghiệm duy nhất \(\left(x;y\right)=\left(4;1\right).\)

Bạn sửa lại đề dùm mình nha, sai đề hơi nhiều đó.

ĐKXĐ:\(x\ne0;2\)

\(P=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2+2x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}-\frac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right).\frac{x^2-x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}+\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right).\frac{x^2-2x+x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)^2}{2\left(x^2+4\right)\left(x-2\right)}+\frac{4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right).\frac{x\left(x-2\right)+\left(x-2\right)}{x^2}\)

\(P=\frac{x\left(x^2-4x+4\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{x^3-4x^2+4x-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{\left(x^3+4x\right)\left(x-2\right)\left(x+1\right)}{2\left(x^2+4\right)\left(x-2\right).x^2}\\ P=\frac{x\left(x^2+4\right)\left(x-2\right)\left(x+1\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\\ P=\frac{x+1}{2x}\)

A=(2x+3)(x^2-x+2) - 2x(x^2+1/2x-3)

=(2x^3- 2x^2+4x+3x^2-3x+6) - (2x^3+x^2-6x)

=2x^3 - 2x^2 + 4x + 3x^2 - 3x + 6 - 2x^3 - x^2 - 6x

=6 - 5x

ta có : (x+2)^2 + (x-3)^2 - 2(x-3)(x+2)=4x+13

<=> [(x+2) -(x-3)]^2=4x+13

<=>(x+2-x+3)=4x+13

<=>4x+13=5

=>4x= 5-13=-8

=>x=-8/4=-2

17x+3.(-16x-37)    =2x+43+4x

17x+3.(-16).x-3.37=6x+43

17x+(-48)x-119     =6x+43

[17x-6x+(-48)x]     =43+119

-37x                       =162

37x                        =-162

x                            =-162:37

x                             = -162/37

A) x² - 4x + 4 = (x - 2)² (hằng đẳng thức số 2)

Cm : x² - 4x + 4 = x² - 2x - 2x + 4 = x(x - 2) - 2(x - 2) = (x - 2)(x - 2) = (x - 2)²

b tương tự