\(a,\Leftrightarrow\dfrac{\left(x-3\right)^2-\left(x+3\right)^2-48}{x^2-9}=0\)

\(\Leftrightarrow x^2-6x+9-x^2-6x-9-48=0\)

\(\Leftrightarrow-12x-48=0\)

\(\Leftrightarrow-12x=48\)

\(\Leftrightarrow x=-4\)

\(b,\Leftrightarrow\dfrac{\left(x-5\right)\left(x+1\right)-\left(2x+3\right)-x\left(x-1\right)}{x^2-1}=0\)

\(\Leftrightarrow x^2+x-5x-5-2x-3-x^2+x=0\)

\(\Leftrightarrow-5x-8=0\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\dfrac{8}{5}\)

Mk giải giúp bạn phần a thôi nha! (Dài lắm, lười :v)

a, 1 + \(\dfrac{x}{3-x}\) = \(\dfrac{5x}{\left(x+2\right)\left(x+3\right)}+\dfrac{2}{x+2}\) (x \(\ne\) -2; x \(\ne\) \(\pm\) 3)

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{5x+2\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}\)

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{5x+2x+6}{\left(x+2\right)\left(x+3\right)}\)

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{7x+6}{x^2+5x+6}\)

Vì 3 - x \(\ne\) 0; x² + 5x + 6 \(\ne\) 0

\(\Rightarrow\) 3(x² + 5x + 6) = (7x + 6)(3 - x)

\(\Leftrightarrow\) 3x² + 15x + 18 = 21x - 7x² + 18 - 6x

\(\Leftrightarrow\) 10x² = 0

\(\Leftrightarrow\) x = 0 (TM)

Vậy S = {0}

Chúc bn học tốt! (Nếu bạn cần phần nào khác mk có thể giúp bn chứ đừng có đăng hết lên, ít người làm lắm :v)

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

vậy..

d)\(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(2-x\right)}+\dfrac{\left(x+3\right)\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}=0\\ \Leftrightarrow5+x^2-9=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

vậy..

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

$\Leftrightarrow (2x+1-x)(2x+1+x)=0$

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

$\Leftrightarrow (x-1)^2-2^2=0$

$\Leftrightarrow (x-1-2)(x-1+2)=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

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

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

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

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

a)

 \(1+\dfrac{x+1}{3}>\dfrac{2x-1}{6}-2\\ \Leftrightarrow6+2\left(x+1\right)>2x-1-12\\ \Leftrightarrow8>-13\left(t.m\right)\)

Vậy bất phương trình có vô số nghiệm.

a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

=>x=0

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

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

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

\(\Leftrightarrow x^2+2x+1=0\)

=>x=-1(nhận)

\(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

\(\Leftrightarrow x=0\) ( vì \(x^2+2x+5>0;\forall x\)

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

\(ĐK:x\ne1;3;4\)

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

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

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

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

\(\Leftrightarrow x^2-3x-x^2+4x-3=x^2-1\)

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

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

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

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

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

Vậy \(S=\left\{-2\right\}\)

\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)

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

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

\(\Rightarrow-x^2-2x-1=0\)

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

a) ĐKXĐ: \(x\notin\left\{-3;2;-1;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)

\(\Leftrightarrow\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{2}{\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x-1}\)

\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}-\dfrac{2\left(x-2\right)}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)

\(\Leftrightarrow\dfrac{5x+5-2x+4}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)

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

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

Suy ra: \(\left(x+1\right)\left(x-2\right)=1-2x\)

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

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

\(\Delta=1^2-4\cdot1\cdot\left(-3\right)=13\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{13}}{2}\left(nhận\right)\\x_2=\dfrac{-1+\sqrt{13}}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{-1-\sqrt{13}}{2};\dfrac{-1+\sqrt{13}}{2}\right\}\)

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

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

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

a: =>-3x=-12

=>x=4

b: =>3(3x+2)-3x-1=12x+10

=>9x+6-3x-1=12x+10

=>12x+10=6x+5

=>6x=-5

=>x=-5/6

c: =>x(x+1)+x(x-3)=4x

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=3(loại) hoặc x=0(nhận)

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

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+5x=30-144\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: S={6}

b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)

\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)

\(\Leftrightarrow2-10x=-12x+12\)

\(\Leftrightarrow2-10x+12x-12=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

hay x=5

Vậy: S={5}

c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow6-2x-8=5x+10\)

\(\Leftrightarrow-2x+2-5x-10=0\)

\(\Leftrightarrow-7x-8=0\)

\(\Leftrightarrow-7x=8\)

hay \(x=-\dfrac{8}{7}\)

Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)

d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)

\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)

\(\Leftrightarrow35-15x-2x-10-10=0\)

\(\Leftrightarrow-17x+15=0\)

\(\Leftrightarrow-17x=-15\)

hay \(x=\dfrac{15}{17}\)

Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)

a) Ta có: ⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30

hay x=6

Vậy: S={6}

b) Ta có: 

Vậy: 7−3x2−5+x5=17−3x2−5+x5=1

Vậy: x+45−x+4=x3−x−22x+45−x+4=x3−x−22

hay x=5

Vậy: S={5}

c) Ta có: ⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4

hay S={−87}S={−87}

d) Ta có: ⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010

hay S={1517}