3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) \(\left|2,5-x\right|-1,3=0\)

th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)

th2: \(2,5-x< 0\Leftrightarrow x>2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)

vậy \(x=1,2;x=3,8\)

b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)

c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)

th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)

th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)

vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)

d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)

th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)

vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)

e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm

a: \(\Leftrightarrow\dfrac{5}{2}:\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{15}{4}-3=\dfrac{3}{4}\)

\(\Leftrightarrow\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{5}{2}:\dfrac{3}{4}=\dfrac{5}{2}\cdot\dfrac{4}{3}=\dfrac{20}{6}=\dfrac{10}{3}\)

=>3/4x+1/2=10/3 hoặc 3/4x+1/2=-10/3

=>3/4x=17/6 hoặc 3/4x=-23/6

=>x=34/9 hoặc x=-46/9

b: \(\Leftrightarrow\dfrac{9}{4}:\left|x+\dfrac{1}{3}\right|=6.5-2=\dfrac{9}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{9}{4}:\dfrac{9}{2}=\dfrac{1}{2}\)

=>x+1/3=1/2 hoặc x+1/3=-1/2

=>x=1/6 hoặc x=-5/6

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

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

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

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

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

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(a,4x-12=0\\ \Leftrightarrow x=3\\ b,x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\\ \Leftrightarrow\left(x^2+x\right)-\left(x^2-x-6\right)-7=0\\ \Leftrightarrow x^2+x-x^2+x+6-7=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+3-x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Rightarrow-4x+3=0\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

a, \(4x=12\Leftrightarrow x=3\)

b, \(x^2+x-x^2+x+6=7\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

c, đk : x khác -1 ; 1 

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

a.

`4x-12=0`

`<=> 4x=12`

`<=> x= 3`

b.

`x(x+1)-(x+2)(x-3)=7`

`<=> (x^2+x)-(x^2-x-6)-7=0`

`<=> x^2+x+x^2+x+6-7=0`

`<=> 2x-1=0`

`<=> x= 1/2`

c. 

`(x-3)/(x+1)= (x^2)/(x^2-1)` `(đkxđ:` \(\ne\) `+-1`

`<=> ((x-3)(x-1))/((x-1)(x+1)) - (x^2)/((x-1)(x+1)) = 0`

`<=> (x^2-4x+3-x^2)/((x-1)(x+1))=0`

`=> x^2-4x+3-x^2=0`

`<=> -4x+3=0`

`<=> x=3/4` `(tmđk)`

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

A/ Ta có số nào nhân với 0 cx = 0 

Vậy từ đó suy ra 2 trường hợp 

TH1\(4x-9=0\)

\(=>x=\frac{9}{4}\)

TH2 \(2,5+-\frac{7}{3}x=0\)

 \(=>x=\frac{15}{14}\)

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

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

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

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

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26