1. \(\left|\frac{2x^2-x}{3x-4}\right|\ge1\) Điều kiện: \(x\ne\frac{4}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2-x}{3x-4}\ge1\\\frac{2x^2-x}{3x-4}\le-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{x^2-2x+2}{3x-4}\ge0\\\frac{x^2+x-2}{3x-4}\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x>\frac{4}{3}\\x\in(-\infty;-2]U[1;\frac{4}{3})\end{cases}}\Leftrightarrow x\in(-\infty;-2]U[1;+\infty)\backslash\left\{\frac{4}{3}\right\}\)

2.\(\hept{\begin{cases}x^2\le-2x+3\left(1\right)\\\left(m+1\right)x\ge2m-1\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow x^2+2x-3\le0\Leftrightarrow-3\le x\le1\)

+) Nếu \(m=-1\) thì (2) vô nghiệm, suy ra \(m\ne-1\)

+) Nếu \(m>-1\) thì \(\left(2\right)\Leftrightarrow x\ge\frac{2m-1}{m+1}\)

Hệ BPT có nghiệm duy nhất \(\Leftrightarrow\frac{2m-1}{m+1}=1\Leftrightarrow m=2>-1\)

+) Nếu \(m< -1\)thì \(\left(2\right)\Leftrightarrow x\le\frac{2m-1}{m+1}\)

Hệ BPT có nghiệm duy nhất \(\Leftrightarrow\frac{2m-1}{m+1}=-3\Leftrightarrow m=-\frac{2}{5}< -1\)

Vậy \(m=\left\{\frac{-2}{5};2\right\}\)

1. |2x2−x3x−4 |≥1 Điều kiện: x≠43 

⇔[

⇔[

⇔[

⇔x∈(−∞;−2]U[1;+∞)\{43 }

2.{

(1)⇔x2+2x−3≤0⇔−3≤x≤1

∣∣∣∣∣​3x−42x2−x​∣∣∣∣∣​≥1⇔⎣⎢⎢⎢⎡​​3x−42x2−x​≥13x−42x2−x​≤−1​⇔⎣⎢⎢⎢⎡​​3x−42x2−4x+4​≥03x−42x2+2x−4​≤0​⇔⎣⎢⎢⎡​​3x−4>03x−42x2+2x−4​≤0​

\Leftrightarrow \left[\begin{aligned}&{x>\dfrac{4}{3} } \\ &{1\le x<\dfrac{4}{3} } \\ &{x\le -2} \end{aligned}\right.⇔⎣⎢⎢⎢⎢⎢⎡​​x>34​1≤x<34​x≤−2​ .

Tập nghiệm :S=\left(-\infty ;-2\right]\cup \left[1;\dfrac{4}{3} \right)\cup \left(\dfrac{4}{3} ;+\infty \right)S=(−∞;−2]∪[1;34​)∪(34​;+∞).

2.

Ta có: \left\{\begin{aligned}&{x^{2} \le -2x+3} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&{x^{2} +2x-3\le 0} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&{-3\le x\le 1} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right.{​x2≤−2x+3(m+1)x≥2m−1​⇔{​x2+2x−3≤0

2.a)

\(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\)

\(\Leftrightarrow-3x>6\)

\(\Leftrightarrow3>\dfrac{6}{-3}\)

\(\Leftrightarrow x< -2\)

Vậy nghiệm của bpt \(S=\left\{-2\right\}\)

2.b)

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

\(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\)

\(\Leftrightarrow4x+4-12\ge3x-6\)

\(\Leftrightarrow4x-3x\ge-6-4+12\)

\(\Leftrightarrow x\ge2\)

vậy nghiệm của bpt x\(\ge\)2

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

7,3, -6

ĐKXĐ: \(x\ne7;x\ne2\)

BPT \(\Leftrightarrow f\left(x\right)=\dfrac{\left(6-2x\right)^3\left(x+6\right)}{\left(x-7\right)^3}\le0\)

Lập bảng xét dấu ta có:

Từ đây ta thấy \(-6\le x\le3\) hoặc \(x>7\) thỏa mãn bất phương trình ban đầu.

Vậy...

a,Áp dụng BĐT `|A|-|B|<=|A-B|`

`=>|x+1|-|x-2|<=|x+1-x+2|=3`

Mà đề bài `|x+1|-|x-2|>=3`

`=>|x+1|-|x-2|=3`

`=>x=2\or\x=-1`

`b,1/(|x|-3)-1/2<0`

`<=>(5-|x|)/(2|x|-6)<0`

`<=>(|x|-5)/(|x|-3)>0`

`<=>` $\left[ \begin{array}{l}|x|>5\\|x|<3\end{array} \right.$

`<=>` $\left[ \begin{array}{l}\left[ \begin{array}{l}x>5\\x<-5\end{array} \right.\\-3<x<3\end{array} \right.$

`-2<=x<=1` nhé câu a ý mình ghi thiếu.

đkxđ: x khác 0

\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)

\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)

\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)

\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)

\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)

\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)

=> x=0 ( loại , ko tm)

Vậy pt vô nghiệm

Phương pháp:

Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)

Thay vào pt

\(x\ne0:đặt:x+\dfrac{1}{x}=t\)

\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)

\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)

\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)

\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)

(bn xem lại đề)