a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

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

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

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

a: ĐKXĐ: x<>-2/3

\(\frac{2x+1}{3x+2}=5\)

=>5(3x+2)=2x+1

=>15x+10=2x+1

=>13x=-9

=>\(x=-\frac{9}{13}\) (nhận)

b: ĐKXĐ: x∉{1;3}

\(\frac{2x^2-5x+2}{x-1}=\frac{2x^2+x+15}{x-3}\)

=>\(\left(2x^2-5x+2\right)\left(x-3\right)=\left(2x^2+x+15\right)\left(x-1\right)\)

=>\(2x^3-6x^2-5x^2+15x+2x-6=2x^3-2x^2+x^2-x+15x-15\)

=>\(-11x^2+17x-6=-x^2+14x-15\)

=>\(-10x^2+3x+9=0\)

=>\(10x^2-3x-9=0\)

=>\(x^2-\frac{3}{10}x-\frac{9}{10}=0\)

=>\(x^2-2\cdot x\cdot\frac{3}{20}+\frac{9}{400}-\frac{9}{400}-\frac{9}{10}=0\)

=>\(\left(x-\frac{3}{20}\right)^2=\frac{9}{400}+\frac{9}{10}=\frac{9}{400}+\frac{360}{400}=\frac{369}{400}\)

=>\(x-\frac{3}{20}=\pm\frac{3\sqrt{41}}{20}\)

=>\(\left[\begin{array}{l}x=\frac{3\sqrt{41}+3}{20}\left(nhận\right)\\ x=\frac{-3\sqrt{41}+3}{20}\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: x∉{3;-3}

\(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{x^2-9}+2\)

=>\(\frac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{24+2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)

=>(2x+3)(x+3)-4(x-3)=\(24+2x^2-18\)

=>\(2x^2+6x+3x+9-4x+12=2x^2+6\)

=>5x+21=6

=>5x=-15

=>x=-3(loại)

\(ĐKXĐ:x\ne-2\) 

Ta thấy x=0 ko là nghiệm của phương trình. Do đó \(x\ne0\)

 \(\Rightarrow\dfrac{1}{\dfrac{x^2+4x+4}{x}}+\dfrac{5}{\dfrac{x^2+4}{x}}=-2\) (chia cả tử và mẫu của 2 phân số vế trái cho x )

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

Đặt \(x+\dfrac{4}{x}=t\) (\(t\ne0,t\ne-4\))

\(pt\) trở thành: \(\dfrac{1}{t+4}+\dfrac{5}{t}=-2\) \(\Rightarrow t+5\left(t+4\right)=-2\left(t+4\right)t\Leftrightarrow t+5t+20=-2t^2-8t\Leftrightarrow2t^2+14t+20=0\Leftrightarrow t^2+7t+10=0\) \(\Leftrightarrow\left(t+2\right)\left(t+5\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-2\left(1\right)\\t=-5\left(2\right)\end{matrix}\right.\)

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

Từ (2) \(\Rightarrow x+\dfrac{4}{x}=-5\Rightarrow x^2+4=-5x\Leftrightarrow x^2+5x+4=0\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-4\left(TM\right)\end{matrix}\right.\) Vậy...

*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)

<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)

=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2

<=>x-2=5x-2

<=>x-5x=2-2

<=>-4x=0

<=> x = 0(TM)

Vậy phương trình có tập nghiệm là S={0}

*(x+4)(5x+9)-x-4=0

<=>(x+4)(5x+9)-(x+4)=0

<=>(x+4)(5x+9-1)=0

<=>(x+4)(5x+8)=0

<=>x+4= 0 hoặc 5x+8=0

(+) x+4=0 (+)5x+8=0

<=>x=-4 <=>5x=-8

<=>x=\(\dfrac{-8}{5}\)

Vậy phương trình có tập nghiệm là S={\(-4;\dfrac{-8}{5}\)}

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

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

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

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

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

\(\Leftrightarrow x=3\)

ĐKXĐ: x∉{2;3;4;5;6}

Sửa đề: \(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac58\)

=>\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac58\)

=>\(-\frac{1}{x-2}+\frac{1}{x-3}-\frac{1}{x-3}+\frac{1}{x-4}-\frac{1}{x-4}+\frac{1}{x-5}-\frac{1}{x-5}+\frac{1}{x-6}=\frac58\)

=>\(\frac{1}{x-6}-\frac{1}{x-2}=\frac58\)

=>\(\frac{x-2-\left(x-6\right)}{\left(x-2\right)\left(x-6\right)}=\frac58\)

=>5(x-2)(x-6)=32

=>\(5\left(x^2-8x+12\right)=32\)

=>\(5x^2-40x+28=0\)

=>\(x^2-8x+\frac{28}{5}=0\)

=>\(x^2-8x+16-10,4=0\)

=>\(\left(x-4\right)^2=10,4\)

=>\(\left[\begin{array}{l}x-4=\sqrt{10,4}\\ x-4=-\sqrt{10,4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt{10,4}+4\left(nhận\right)\\ x=-\sqrt{10,4}+4\left(nhận\right)\end{array}\right.\)

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

⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

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

⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)

⇔ \(x^2+8x+12=32\)

⇔ \(x^2+8x-20=0\)

⇔ \(\left(x-2\right)\left(x+10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

Sửa lại đề nha:

 \(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)

a.

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)

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

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)