giải pt sau

g) 11+8x-3=5x-3+x

\(\Leftrightarrow\) 8x + 8 = 6x - 3

<=> 8x-6x = -3 - 8

<=> 2x = -11

=> x=-\(\dfrac{11}{2}\)

Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}

h)4-2x+15=9x+4-2x

<=> 19 - 2x = 7x + 4

<=> -2x - 7x = 4 - 19

<=> -9x = -15

=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)

Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}

g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)

<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)

<=> 9x + 6 - 3x + 1 = 10 + 12x

<=> 6x + 7 = 10 + 12x

<=> 6x -12x = 10-7

<=> -6x = 3

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

Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}

\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)

<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)

<=> x + 4 - 5x - 20 = 4x + 2 - 25

<=> x - 5x - 4x = 2-25-4+20

<=> -8x = -7

=> x= \(\dfrac{7}{8}\)

Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}

\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)

<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)

<=> 84x + 63 - 90x + 30 = 175x + 140 + 315

<=> 84x - 90x - 175x = 140 + 315 - 63 - 30

<=> -181x = 362

=> x = -2

Vậy tập nghiệm của PT là : S={-2}

K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)

<=> 25x + 10 - 80x - 10 = 24x + 12 - 150

<=> -55x = 24x - 138

<=> -55x - 24x = -138

=> -79x = -138

=> x=\(\dfrac{138}{79}\)

Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}

m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

<=> 6x - 3 - 5x + 10 = x+7

<=> x + 7 = x+7

<=> 0x = 0

=> PT vô nghiệm

Vậy S=\(\varnothing\)

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

<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)

<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)

<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)

=> x= 1

Vậy S={1}

p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)

<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)

<=> 2x -2x + 1= x-36

<=> 2x-2x-x = -37

=> x = 37

Vậy S={37}

q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)

<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)

<=> 8 + 4x - 10x = 5 - 10x + 5

<=> 4x-10x + 10x = 5+5-8

<=> 4x = 2

=> x= \(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}

g) \(11+8x-3=5x-3+x\)

\(\Leftrightarrow8+8x=6x-3\)

\(\Leftrightarrow8x-6x=-3-8\)

\(\Leftrightarrow2x=-11\)

\(\Leftrightarrow x=-\dfrac{11}{2}\)

h, \(4-2x+15=9x+4-2x\)

\(\Leftrightarrow-2x-9x+2x=4-4-15\)

\(\Leftrightarrow-9x=-15\)

\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)

g) 11+8x-3=5x-3+x

=> 8x -5x -x = -3 -11+3

<=> 2x = -11

<=> x = \(\dfrac{-11}{2}\)

h)4-2x+15=9x+4-2x

=> -2x -9x +2x = 4-4-15

<=> -9x = -15

<=> x = \(\dfrac{5}{3}\)

a: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

hay x=1/7

b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

=>12x+10=6x+5

=>6x=-5

hay x=-5/6

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

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

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)

ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)

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

vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn

bài này đề bài là chứng minh hay là giải bất phương trình vậy bạn

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

Mà \(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

c)

\(\left(x^2-2x\right)\left(x^3-3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^3-6x^2+3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left[\left(x^3-6x^2\right)+\left(3x^2-18x\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left[x^2\left(x-6\right)+3x\left(x-6\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-6\right)\left(x^2+3x\right)=0\)

\(\Leftrightarrow x^2\left(x-6\right)\left(x-2\right)\left(x+3\right)=0\)

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

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)