bài nào ạ

À vậy  để c đăng lại nha.

ĐK:...

\(\Leftrightarrow\dfrac{2x+3}{x-3}-\dfrac{4}{x+3}-\dfrac{24}{\left(x-3\right)\left(x+3\right)}-2=0\)

\(\Leftrightarrow\dfrac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)-24-2\left(x^2-9\right)}{x^2-9}=0\)

\(\Leftrightarrow2x^2+6x+3x+9-4x+12-24-2x^2+18=0\)

\(\Leftrightarrow5x+15=0\)

<=> x = -3 (ko t/m đk)

=> Pt vô nghiệm

bạn ktra lại đề nhé 

ĐK: `x \ne 0 ;x \ne -4`

`24/(x+4)-24/x=1/2`

`<=>24 ( 1/(x+4) -1/x) = 1/2`

`<=> 1/(x+4)-1/x=1/48`

`<=> x-(x+4)=48x(x+4)`

`<=>-4=48x^2+192`

`<=>48x^2+196=0` (VN)

Vậy không có `x` thỏa mãn.

\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\left(x\text{ ≥}1\right)\)

⇔ \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

⇔ \(-\sqrt{x-1}=-17\)

⇔ \(x=290\left(TM\right)\)

KL..................

a,ĐK: x\(\ge\)1

⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)

⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

\(a.x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\) ( ĐKXĐ : \(x\ne0\) )

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

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

\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\\dfrac{1}{x^2}-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\1-x^3=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\Leftrightarrow x=1\) ( x² + x + 1 loại nhé nếu phân tích ra thì ta được \(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\) )

Vậy \(S=\left\{1\right\}\)

b, \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow x\left(x+3\right).\left(x+1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x+1\right)-1-24=0\Leftrightarrow\left(x^2+3x+1\right)-25=0\)

\(\Leftrightarrow\left(x^2+3x+1-5\right)\left(x^2+3x+1+5\right)=0\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{-4;1\right\}\)

e, \(\left(x^2+x+1\right)-2x^2-2x=5\Leftrightarrow\left(x^2+x+1\right)-2x^2-2x-2-3=0\)

\(\Leftrightarrow\left(x^2+x+1\right)-2\left(x^2+x+1\right)-3=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x-1\right)-3=0< =>\left(x^2+x\right)^2-4=0\) 

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

\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\) ( x^2 + x + 2 loại nhé y như mấy câu trên luôn khác 0 ! )

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

Vậy \(S=\left\{-2;1\right\}\)

\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

d. ĐKXĐ: $x>\frac{-2}{3}$

PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \frac{1}{3x+2}=1$

$\Leftrightarrow 3x+2=1$

$\Leftrightarrow x=-\frac{1}{3}$

Đề có vẻ sai sai. Bạn xem lại đề xem có đúng không?