|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

$\Leftrightarrow (2x+1-x)(2x+1+x)=0$

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

$\Leftrightarrow (x-1)^2-2^2=0$

$\Leftrightarrow (x-1-2)(x-1+2)=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

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

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

a)\(=>\left[{}\begin{matrix}3x-2=0\\2x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}3x=2\\2x=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(ĐK:x\ne\pm1\)

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

\(\Leftrightarrow\left(x-1\right)^2-4=2\left(x+1\right)^2\)

\(\Leftrightarrow x^2-2x+1-4=2x^2+4x+2\)

\(\Leftrightarrow x^2+6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

=>[3x−2=02x+5=0=>[3x=22x=−5=>⎡⎢ ⎢⎣x=23x=−52

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

a) \(5x-3=7\)

\(\Leftrightarrow5x=7+3\)

\(\Leftrightarrow5x=10\)

\(\Leftrightarrow x=\dfrac{10}{5}\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

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

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

*) \(x+3=0\)

\(x=0-3\)

\(x=-3\)

*) \(x-4=0\)

\(x=0+4\)

\(x=4\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\left|x^2+2014\right|=1\)

\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)

*) \(x^2+2014=1\)

\(\Leftrightarrow x^2=1-2014\)

\(\Leftrightarrow x^2=-2013\) (vô lý)

*) \(x^2+2014=-1\)

\(\Leftrightarrow x^2=-1-2014\)

\(\Leftrightarrow x^2=-2015\) (vô lý)

Vậy \(S=\varnothing\)

d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)

ĐKXĐ: \(x\ne-1;x\ne3\)

\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-6-x-1=3x-11\)

\(\Leftrightarrow-2x=-11+7\)

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

\(\Leftrightarrow x=2\) (nhận)

Vậy \(S=\left\{2\right\}\)

2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)

<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)

<=>9x-6=2-4x

<=>9x+4x=2+6

<=>13x=8

<=>x=\(\dfrac{8}{13}\)

1.a)2(x-0,5)+3=0,25(4x-1)

<=>2x-1+3=x-1phần4

<=>2x-x=-1/4+1-3

<=>x=-3/4

a. 2( x - 0,5 ) + 3 = 0,25(4x - 1 )

(=) 2x - 1 + 3 = x - 0,25

(=) 2x - x = 1 - 3 - 0,25

(=) x = -2,25

b. 2.

(=) 2x - 2x = 0,5 + 4 -3 +2 ( vô lí )

Vậy phương trình này vô nghiệm

a: =>(x-2)(2x+5)=0

=>x-2=0 hoặc 2x+5=0

=>x=2 hoặc x=-5/2

c: \(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)

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

=>x^2+3x=x^2-1

=>3x=-1

=>x=-1/3

\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)

\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\)              ( ĐKXĐ: \(x\ne-1;x\ne1\) )

\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\)  ( nhận )

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

\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)

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