a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)

\(\Leftrightarrow x+1-\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

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

=>3x=5x+12

=>-2x=12

hay x=-6(nhận)

\(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)\(\Leftrightarrow\)\(\dfrac{x^2+2x+1+1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)

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

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

\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = x + 2 + x + 3 - x - 1 - x - 4

\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = 0

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

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

\(\Leftrightarrow\) \(\dfrac{x+4+4x+4}{x^2+5x+4}\) = \(\dfrac{2x+6+3x+6}{x^2+5x+6}\)

\(\Leftrightarrow\) \(\dfrac{5x+8}{x^2+5x+4}\) = \(\dfrac{5x+12}{x^2+5x+6}\)

Đặt 5x + 8 = y; x² + 5x + 4 = t, ta có:

\(\dfrac{y}{t}\) = \(\dfrac{y+4}{t+2}\)

\(\Leftrightarrow\) \(\dfrac{y\left(t+2\right)}{t\left(t+2\right)}\) = \(\dfrac{t\left(y+4\right)}{t\left(t+2\right)}\)

\(\Leftrightarrow\) yt + 2y = yt + 4t

\(\Leftrightarrow\) 2y = 4t

\(\Leftrightarrow\) 2(5x + 8) = 4(x² + 5x + 4)

\(\Leftrightarrow\) 10x + 16 = 4x² + 20x + 16

\(\Leftrightarrow\) 16 - 16 = 4x² + 20x - 10x

\(\Leftrightarrow\) 0 = 4x² + 10x

\(\Leftrightarrow\) 2x(2x + 5) = 0

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

CHÚC BN HOK TỐT...

a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)

=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)

=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22

=>-x^2+59x+14-8x^2+5x+22=0

=>-9x^2+54x+36=0

=>x^2-6x-4=0

=>\(x=3\pm\sqrt{13}\)

b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)

=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)

=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32

=>x^2+6x+19=x^2+4x-32

=>2x=-51

=>x=-51/2

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

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

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)

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

\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

\(a,\left(đk:x\ge0\right)\) 

\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)

\(x>0\)

\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)

\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)

\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)

\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)

a) ĐKXĐ : \(x\ge0\)

PT <=> \(x+3-4\sqrt{x}\sqrt{x+3}+4x=0\)

<=> \(\left(\sqrt{x+3}-2\sqrt{x}\right)^2=0\)

<=> \(\sqrt{x+3}=2\sqrt{x}\)

<=> \(x+3=4x\)

<=> x = 1

Vậy x = 1 là nghiệm phương trình