`8(x-3)(x+1)=8x^2 +11`

`<=>8(x^2 +x-3x-3)-8x^2 -11=0`

`<=>8x^2 +8x-24x-24-8x^2 -11=0`

`<=>-16x-35=0`

`<=>-16x=35`

`<=>x=-35/16`

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(x\ne0;x\ne2\right)\\ < =>\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

suy ra

`x^2 +2x-2=x-2`

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

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

`<=>x(x+1)=0`

\(< =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ < =>x=-1\)

\(a,8\left(x-3\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow\left(8x-24\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow8x^2-24x+8x-24-8x^2-11=0\\ \Leftrightarrow-16x-35=0\\ \Leftrightarrow x=\dfrac{-35}{16}\)

Vậy \(x=-\dfrac{35}{16}\)

\(b,đkxđ:x\ne2;x\ne0\)

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}-\dfrac{1}{x}=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=0\\ \Leftrightarrow x^2+2x-2-x+2=0\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)

Vậy \(x=-1\)

@ducminh 

a. ĐKXĐ: \(x\ne2\).

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

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

⇔\(1+3x-6=3-x\)

⇔\(4x-8=0\)

⇔\(x=2\) (không thỏa mãn)

-Vậy S=∅.

b. ĐKXĐ: \(x\ne-1\)

 \(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)

⇔\(\dfrac{5x}{2\left(x+1\right)}+1=-\dfrac{6}{x+1}\)

⇔\(\dfrac{5x}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}=-\dfrac{12}{2\left(x+1\right)}\)

⇔\(5x+2\left(x+1\right)=-12\)

⇔\(5x+2x+2+12=0\)

⇔\(7x+14=0\)

⇔\(x=-2\) (thỏa mãn).

-Vậy \(S=\left\{-2\right\}\)

a, \(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3.\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\\ \Leftrightarrow1+3x-6=3-x\)

\(\Leftrightarrow3x+x=3-1+6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=\dfrac{8}{4}=2\\ Vậy.S=\left\{2\right\}\)

b,  \(\Leftrightarrow\)\(\dfrac{5x}{2x+2}+\dfrac{2x+2}{2x+2}=\dfrac{-6.2}{2.\left(x+1\right)}\)

\(\Leftrightarrow5x+2x+2=-12\\ \Leftrightarrow7x=-12-2\\ \Leftrightarrow7x=-14\\ \Leftrightarrow x=-\dfrac{14}{7}=-2\\ Vậy.S=\left\{-2\right\}\)

a.

ĐKXĐ: \(x>0\)

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

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)

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

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)

\(x+2+x\sqrt{2x+1}=x\sqrt{x+2}+\sqrt{\left(x+2\right)\left(2x+1\right)}\)

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)

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

b.

\(\left(x^2+1\right)^2=5-x\sqrt{2x^2+4x}\)

\(\Leftrightarrow x^4+2x^2-4+x\sqrt{2x^2+4x}=0\)

Đặt \(x\sqrt{2x^2+4x}=t\Rightarrow t^2=x^2\left(2x^2+4x\right)=2\left(x^4+2x^2\right)\)

Pt trở thành:

\(\dfrac{t^2}{2}-4+t=0\)

\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4x}=2\left(x>0\right)\\x\sqrt{2x^2+4x}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\left(x>0\right)\\x^4+2x^2-8=0\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\Rightarrow\dfrac{2x^2+9}{x^2}=\dfrac{1}{t^2}\)

Pt trở thành:

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

\(\Rightarrow2t^3-3t^2+1=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(2t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vô-nghiệm\right)\\4x^2=2x^2+9\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{3\sqrt{2}}{2}\)

Kiểm tra lại vế trái đề bài câu b

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)

\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)

\(\Leftrightarrow4-2x-6x-12\le3x-51\)

\(\Leftrightarrow-11x\le-43\)

\(\Leftrightarrow x\ge\dfrac{43}{11}\)

Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)

\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)

\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)

\(\Leftrightarrow0x\le-10\) (vô lý)

Vậy \(S=\varnothing\)

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow-x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

a, đk : x khác 5;-6 

\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm) 

b, đk : x khác 1;3 

\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)

pt vô nghiệm 

a, đk : x khác 5;-6 

(tm) 

b, đk : x khác 1;3 

x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)

pt vô nghiệm 

a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>x=0(nhận)

b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=-6

hay x=-3(nhận)