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: \(S=\left\{-\dfrac{3}{7}\right\}\)

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

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

Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

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

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

1)

<=> \(x^2-3x=0\)

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

x= 0 

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

\(2,x^2-3x=4\)

\(\Leftrightarrow x^2-3x-4=0\)

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

\(4,x^3=9x\)

\(\Leftrightarrow x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

Vậy \(S=\left\{0;\pm3\right\}\)

\(5,\left(x+2\right)\left(x-3\right)=x^2-4\)

\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)

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

\(\Leftrightarrow-x=2\)

\(\Leftrightarrow x=-2\)

Vậy \(S=\left\{-2\right\}\)

\(ĐKXĐ:x\ne-2\) 

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...

a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)

\(\Leftrightarrow15x-9x+6=45-10x+25\)

\(\Leftrightarrow15x-9x+10x=45+25-6\)

\(\Leftrightarrow16x=64\)

\(\Leftrightarrow x=4\)

b) \(x^2-9+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

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

\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)

\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)

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

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

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

a) 15x - 3(3x - 2) = 45 - 5(2x - 5)

\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25

\(\Leftrightarrow\) 6x + 10x = 70 - 6

\(\Leftrightarrow\) 16x = 64

\(\Leftrightarrow\) x = 4

Vậy.......................

b) x² - 9 + 4(x - 3) = 0

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

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

\(\Leftrightarrow\) (x - 3)(x + 7) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)

Vậy........................

c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)

\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)

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

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

\(\Leftrightarrow\) x² - x - 5x - 4 + 4 = 0

\(\Leftrightarrow\) x² - 6x = 0

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

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

Vậy...............

M nên ở nhà chứ học làm gì nữa bài quá easy

2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)

\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)

Đặt \(x^2+5x+3=t\)

\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)

\(\Leftrightarrow t^2-9=280\)

\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)

\(\Leftrightarrow x^2-2x+7x-14=0\)

\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0

\(\Leftrightarrow\) x = 2 hoặc x = - 7

Vậy x = 2 hoặc x = -7.

3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)

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

\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)

\(\Leftrightarrow x^3+12x^2+46x+60=0\)

\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)

\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)

\(\Leftrightarrow x=-6\)

Vậy x = -6.

4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow2\left[\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right]=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{2}{9}\)

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

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{6}{x\left(x+6\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2x\left(x+6\right)=54\)

\(\Leftrightarrow2x^2+12x-54=0\)

\(\Leftrightarrow2x^2-6x+18x-54=0\)

\(\Leftrightarrow2x\left(x-3\right)+18\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+18\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x+9\right)=0\)

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 9 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -9

Vậy x = 3 hoặc x = -9.

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 \(x=-1\)

d.

ĐKXĐ: \(x>1\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)

\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)

\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow1-\dfrac{1}{ab}=0\)

\(\Leftrightarrow ab=1\)

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

\(\Leftrightarrow x^3-1=1\)

\(\Leftrightarrow x=\sqrt[3]{2}\)