Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

| 2-4x | = 4x-2

<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x-4x+2=0\\2-4x-4x+2=0\end{cases}}\)

<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)

<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)

=> \(S=\left\{\frac{1}{2};\infty\right\}\)

2x-7> 3(x-1)

<=>2x-7>3x-3

<=>2x-3x>-3+7

<=>-x>4

<=>x<4

=>S={x/x<4}

1-2x<4(3x-2)

<=>1-2x<12x-8

<=>-2x-12x<-8-1

<=>-14x<-9

<=>x>\(\frac{9}{14}\)

=>S={\(\frac{9}{14}\)}

-3x+2|-4 -x|> 0

<=>\(\orbr{\begin{cases}-3x+2+4+x>0\\-3x+2-4x-x>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)

=>S={x/x<3;x/x<\(\frac{1}{4}\)}

4x-1|x-2|< 0

<=>\(\orbr{\begin{cases}4x-1-x+2< 0\\4x-1+x-2< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x+1< 0\\3x-3< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)

<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)

=>S={x/x<\(\frac{-1}{3}\);x/x<1}

e, ĐK: \(x\ne2\)

\(\dfrac{3}{x-2}>1\Leftrightarrow\dfrac{5-x}{x-2}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}5-x>0\\x-2>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}5-x< 0\\x-2< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow2< x< 5\)

\(\left(2\right)\Leftrightarrow\) vô nghiệm

Vậy \(2< x< 5\)

f, ĐK: \(x\ne\dfrac{1}{2}\)

\(\dfrac{2x^2+x}{1-2x}\ge1-x\)

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

\(\Leftrightarrow\dfrac{4x-1}{\left(1-2x\right)\left(x-1\right)}\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-1\ge0\\\left(1-2x\right)\left(x-1\right)>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}4x-1\le0\\\left(1-2x\right)\left(x-1\right)< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{4}\\\dfrac{1}{2}< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 1\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left[{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{4}\)

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........................

a) |x-7|=2x+3  (1)

Ta có:|x-7|=x-7<=>x-7 \(\ge\) 0<=>x\(\ge\)7

         |x-7|=-(x-7)<=>x-7<0<=>x<7

Nếu x\(\ge\)  7thì (1) <=>x-7=2x+3

                         <=>x-2x=7+3

                         <=>-x    =  10

                        <=>x=-10 (ko thỏa mãn đk)

Nếu x<7 thì (1) <=>-(x-7)=2x+3

                          <=>-x+7=2x+3

                        <=>-x-2x=-7+3

                        <=>-3x=-4

                       <=>x=4/3 (thỏa mãn đk)

Giúp vs ạ mk đag cần

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4x2 hay là 4x² vậy bạn ?