\(=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\)

ĐK: ` x\ne \pm 3`

`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`

`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`

`<=>x^2+4x+3+x^2-4x+3=x+6`

`<=>2x^2+6=x+6`

`<=>2x^2-x=0`

`<=>x(2x-1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy `S={0; 1/2}`.

ĐKXĐ: x ≠ -3, x ≠ 3

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)

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

\(\Leftrightarrow2x^2-x=0\)

\(\Leftrightarrow x\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

Vậy...

\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)(a)

ĐKXĐ\(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+1\right).\left(x+3\right)+\left(x-1\right).\left(x-3\right)=x+6\)

\(\Leftrightarrow x^2+3x+x+3+x^2-3x-x+3=x+6\)

\(\Leftrightarrow x^2+3x+x+x^2-3x-x-x=6-3-3\)

\(\Leftrightarrow2x^2-x=0\)

\(\Leftrightarrow x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(thỏa-mãn-ĐKXĐ\right)\\x=\dfrac{1}{2}\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{0;\dfrac{1}{2}\right\}\)

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=> 21x+14<=3x-12`

`<=>18x <= -26`

`<=> x <=-13/9`

\(\dfrac{3-3x}{5}=\dfrac{x-1}{2}\\ \Leftrightarrow2\left(3-3x\right)=5x-5\\ \Leftrightarrow6-6x=5x-5\\ \Leftrightarrow-6x+5x=-5-6\\ \Leftrightarrow-x=-11\\ \Rightarrow x=11\)

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=>21x+14 <= 3x-12`

`<=> 18x <=-26`

`<=>x <= -13/9`

Vậy `x<=-13/9`.

Mik thấy ở vế đầu tiên nó hình như bạn bị nhầm thì  phải :\(\dfrac{2X}{2X+Y}-\dfrac{3Y}{2X-Y}=1\)

Đáp án:x+2/2x

Ta có:2x+6=2(x+3);x²+3x=x(x+3)

➞MTC:2x(x+3)

Ta co:(x+1/2x+6)+(2x+3/x²+3x)={[(x+1)x]+[(2x+3)2]}/2x(x+3)=x²+5x+6/2x(x+3)=(x+2)(x+3)/2x(x+3)=x+2/2x

Ta có: \(\left(\dfrac{1}{3}-2x\right)^{2018}\ge0\forall x\);

\(\left(3y-x\right)^{2020}\ge0\forall x;y\)

=> \(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\ge0\)

mà theo đề thì:\(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\le0\)

=> Dấu ''='' xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\)

Ta có: \(\dfrac{1}{3}-2x=0\Rightarrow x=\dfrac{1}{6}\);

\(3y-x=0\Leftrightarrow3y-\dfrac{1}{6}=0\Leftrightarrow3y=\dfrac{1}{6}\Leftrightarrow y=\dfrac{1}{18}\)

=> \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\dfrac{1}{6}}+\dfrac{1}{\dfrac{1}{18}}=6+18=24\left(đpcm\right)\)