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.....

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

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,ĐK: x\(\ge\)1

⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)

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

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\)

\(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\)

Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\)

\(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \)

Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\)

a ) \(\left|x+5\right|=3x+1\) ( 1 )

+ ) \(x+5=x+5.\) Khi \(x\ge-5\)

\(\left(1\right)\Leftrightarrow x+5=3x+1\)

\(\Leftrightarrow-2x=-4\Leftrightarrow x=2\) ( TM )

+ ) \(x+5=-x-5.\) Khi \(x< -5\)

\(\left(1\right)\Leftrightarrow-x-5=3x+1\)

\(\Leftrightarrow-4x=6\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)( KTM )

Vậy ..........

b ) \(\dfrac{3\left(x-1\right)}{4}+1\ge\dfrac{x+2}{3}\)

\(\Leftrightarrow9x-9+12\ge4x+8\)

\(\Leftrightarrow5x\ge5\)

\(\Leftrightarrow x\ge1\)

Vậy ...........

c ) \(\dfrac{x-2}{x+1}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(1\right)\)

ĐKXĐ : \(x\ne2;x\ne-2.\)

\(\left(1\right)\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\left(x-2\right)^2-3\left(x+2\right)=2x-22\)

\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)

\(\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(TMĐKXĐ\right)\)

Vậy .........

\(\Leftrightarrow\)