1) ĐKXĐ: \(\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

ta có: (-6).\(\sqrt{6x^2-18x+12}\) > \(6x^2-18x-60\)

⇔ \(6x^2-18x+12\) + \(2.3.\sqrt{6x^2-18x+12}+9-81\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+3\right)^2-9^2\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+12\right).\left(\sqrt{6x^2-18x+12}-6\right)\) > 0

⇔ \(\sqrt{6x^2-18x+12}-6\) > 0

⇔ \(\sqrt{6x^2-18x+12}>6\)

⇔\(6x^2-18x+12>36\)

⇔ \(6x^2-18x-24>0\)

⇔\(\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\)

đối chiếu ĐKXĐ ban đầu ta được: x ϵ (-∞;-1) \(\cup\)(4;+∞)

b) ĐKXĐ: \(\forall x\) ϵ R

\(\left(x-2\right)\sqrt{x^2+4}-\left(x-2\right)\left(x+2\right)\le0\)

⇔\(\left(x-2\right)\left(\sqrt{x^2+4}-x-2\right)\le0\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\\sqrt{x^2+4}-x-2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\\sqrt{x^2+4}-x-2\ge0\end{matrix}\right.\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x^2+4\le x^2+4x+4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x^2+4\ge x^2+4x+4\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)

Đối chiếu ĐKXĐ ta được x ϵ ( -∞;0) \(\cup\)( 2; +∞)

a, \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\left(ĐKXĐ:x\ne\pm2;\pm5\right)\)

\(\frac{x+9}{\left(x-5\right)\left(x+2\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}=\frac{1}{x+2}\)

\(\frac{\left(x+9\right)\left(x+5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}=\frac{\left(x+5\right)\left(x-5\right)}{\left(x+2\right)\left(x+5\right)\left(x-5\right)}\)

Khử mẫu : \(\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)=\left(x+5\right)\left(x-5\right)\)

\(x^2+14x+45-x^2-17x-30=x^2-25\)

\(-3x+15-x^2+25=0\)

\(-3x-x^2+40=0\)( giải delta ta đc )

\(x_1=-5;x_2=8\)

b, \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1ĐKXĐ\left(x\ne1;\frac{1}{3}\right)\)

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

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

Khửi mẫu \(x-1+\left(2x+2\right)\left(3x-1\right)-3x^2-1=\left(3x-1\right)\left(x-1\right)\)( bn tự nốt nhé)

c, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)

\(x^2+6x+9-10\ge x^2+5x+6-4\)

\(x-3\ge0\Leftrightarrow x\ge3\)

a) \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\); ĐKXĐ: x # -2; x # +-5

<=> \(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+2}\)

<=> \(\frac{\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}\)

<=> (x + 9)(x + 5) - (x + 15)(x + 2) = (x - 5)(x + 5)

<=> -3x + 15 = x^2 - 25

<=> -3x + 15 - x^2 + 25 = 0

<=> -3x + 40 - x^2 = 0

<=> x^2 + 3x - 40 = 0

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

<=> x - 5 = 0 hoặc x + 8 = 0

<=> x = 5 (ktm0 hoặc x = -8 (tm)

b) \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1\); ĐKXĐ: x # 1/3; x # 1

<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{x\left(3x-1\right)-\left(3x-1\right)}=1\)

<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=1\)

<=> \(\frac{x-1}{\left(x-1\right)\left(3x-1\right)}+\frac{2\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}\)

<=> x - 1 + 2(x + 1)(3x - 1) - 3x^2 + 1 = (x - 1)(3x - 1)

<=> 5x - 4 + 3x^2 = 3x^2 - 4x + 1

<=> 5x - 4 = -4x + 1

<=> 5x + 4x = 1 + 4

<=> 9x = 5

<=> x = 5/9 (tm)

c) (x + 3)^2 - 10 >= (x + 3)(x + 2) - 4

<=> x^2 + 3x + 3x + 9 - 10 >=  x^2 + 2x + 3x + 6 - 4

<=> x^2 + 6x + 9 - 10 >= x^2 + 5x + 6 - 4

<=> x^2 + 6x - 1 >= x^2 + 5x + 2

<=> x^2 + 6x - 1 - x^2 - 5x - 2 >= 0

<=> x - 3 >= 0

<=> x >= 3

2x-x(3x+1)≤15-3x(x+2)

2x-3x²-x≤15-3x² -6x

2x-3x²-x+3x² +6x≤15

7x≤15

x≤15/7

ĐKXĐ: \(x^2+x-1\ge0\)

\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2+b^2>3ab\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)

\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)

\(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0.\)

Vậy \(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0\) khi \(x\in\left(-\infty;-1\right)\cup\left(\dfrac{2}{3};3\right)\cup\left(3;+\infty\right).\)

ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)

\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)

\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)

Vậy ...

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

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

\(\cdotĐKXĐ:x-1\ne0\Leftrightarrow x\ne1 \)
\(x+1\ne0\Leftrightarrow x\ne-1\)

pt: x² + x + x + 1 +3x² - 3x = 4x² + 4x - 4x -4

\(\Leftrightarrow\) x² + 3x² - 4x² + x + x - 3x + 4x - 4x = -4 -1

\(\Leftrightarrow\) - 1x = - 5

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

\(\Leftrightarrow\) x = 5 ( nhận )

Vậy pt có tập nghiệm S= \(\left\{5\right\}\)

2. \(\left|x+2\right|< 2x+10\)

Vì x + 2 < 2x + 10⁽¹⁾ nên x + 2 > 0

-(x + 2) < 2x + 10⁽²⁾ nên - (x + 2) <0

pt(1): x + 2 < 2x + 10

\(\Leftrightarrow\) x - 2x < 10 -2

\(\Leftrightarrow\) -x < 8

\(\Leftrightarrow\) x > -8 ( nhận )

pt(2): -(x + 2) < 2x + 10

\(\Leftrightarrow\) - x - 2 < 2x + 10

\(\Leftrightarrow\) - x - 2x < 10 + 2

\(\Leftrightarrow\) -3x < 12

\(\Leftrightarrow\) x < \(\dfrac{12}{-3}\)

\(\Leftrightarrow\) x < -4 ( nhận)

Vậy bpt có tập nghiệm S= \(\left\{x\left|x< -4\right|\right\}\)

\(\left\{x\left|x>-8\right|\right\}\)

Bài 1.

\(\dfrac{x+1}{x-1}+\dfrac{3x}{x+1}=4\)(đkxđ: x\(\ne\)\(\pm\)

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

\(\Leftrightarrow\) x² + 2x + 1 + 3x² - 3x = 4(x² - 1)

\(\Leftrightarrow\) 4x² - x + 1 = 4x² - 4

\(\Leftrightarrow\) 4x² - 4x² - x = -1 - 4

\(\Leftrightarrow\) -x = -5

\(\Leftrightarrow\) x = 5 (tmđk)

Vậy................

Bài 2.

\(\left|x+2\right|< 2x+10\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x-10< x+2\\x+2>2x+10\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x-x< 10+2\\x-2x>10-2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-3x< 12\\-x>8\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>4\\x< 8\end{matrix}\right.\)

\(\Leftrightarrow\) 4 < x < 8

Vậy........................