a)

\(2x-1+5\left(3-x\right)>0\\ 2x-2+15-5x>0\\ -3x+13>0\\ x< \dfrac{13}{3}.\)

a: =>5(2-x)<3(3-2x)

=>10-5x<9-6x

=>x<-1

b: =>2/9x+5/3>=1/5x-1/5+1/3x

=>2/9x+5/3>=8/15x-1/5

=>-14/45x>=-28/15

=>x<=6

a, \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2;x=3;x=4\)

Vậy tập nghiệm phương trình là S = { 1 ; 2 ; 3 ; 4 } 

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

Vậy tập nghiệm của phương trình là S = { -2 ; 3 } 

c, \(\left|2x-1\right|=\left|x+5\right|\Leftrightarrow\left(2x-1\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(2x-1\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-1-x-5\right)\left(2x-1+x+5\right)=0\Leftrightarrow x=6;x=-\dfrac{4}{3}\)

Vậy tập nghiệm của phương trình là S = { -4/3 ; 6 } 

d, \(\left|3x+1\right|=x-2\)

TH1 : \(3x+1=x-2\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)

TH2 : \(3x+1=-x+2\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

Vậy tập nghiệm của phương trình là S = { -3/2 ; 1/4 } 

các ý còn lại tương tự 

a) Ta có: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)

Vậy: S={3;4;1;2}

b) Ta có: |2x-1|=5

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: S={3;-2}

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

a: ĐKXĐ: \(x\notin\left\{3;-5\right\}\)

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

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

=>\(\dfrac{5x+25-3x+9}{15}=\dfrac{5x+25-3x+9}{\left(x-3\right)\left(x+5\right)}\)

=>(x-3)(x+5)=15

=>\(x^2+2x-15-15=0\)

=>\(x^2+2x-30=0\)

=>\(\left(x+1\right)^2=31\)

=>\(\left[{}\begin{matrix}x+1=\sqrt{31}\\x+1=-\sqrt{31}\end{matrix}\right.\Leftrightarrow x=-1\pm\sqrt{31}\left(nhận\right)\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2+x+1}=3-x\)

=>\(\left\{{}\begin{matrix}x^2+x+1=\left(3-x\right)^2\\x< =3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\x^2-6x+9=x^2+x+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\-7x=-8\end{matrix}\right.\Leftrightarrow x=\dfrac{8}{7}\left(nhận\right)\)

c:

ĐKXĐ: \(x\in R\)

 \(x^2-x+\sqrt{x^2-x+24}=18\)

=>\(x^2-x+24+\sqrt{x^2-x+24}=42\)

=>\(\left(\sqrt{x^2-x+24}\right)^2+\left(\sqrt{x^2-x+24}\right)-42=0\)

=>\(\left(\sqrt{x^2-x+24}+7\right)\left(\sqrt{x^2-x+24}-6\right)=0\)

=>\(\sqrt{x^2-x+24}-6=0\)

=>\(x^2-x+24=36\)

=>\(x^2-x-12=0\)

=>(x-4)(x+3)=0

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

A 3x-4x=-9-3

    -x=-12

     x=12

B 3.2x -5x +1=5+0.2x

  3.2x-5x-0.2x=5-1

  -2x=4

 x=-2

C 1.5-x-2=-3x-0.3

  -x+3x=-0.3-1.5+2

  2x =0.2

  x=0.1

E 2/3-1/2x-1=-x+1

  -1/2x+x=1+1-2/3

  1/2x=4/3

  x=8/3

F 3t-4+13+2t+4-3t

  =3t+2t-3t-4+13+4

  =2t+13

a.

\(\Leftrightarrow3x^3+3x^2+3x=-1\)

\(\Leftrightarrow x^3+3x^2+3x+1=-2x^3\)

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

\(\Leftrightarrow x+1=-\sqrt[3]{2}x\)

\(\Leftrightarrow\left(1+\sqrt[3]{2}\right)x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{1+\sqrt[3]{2}}\)

b.

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

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

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

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

b) Ta có: \(x^3+x^2-x+2=0\)

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

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

\(\Leftrightarrow x+2=0\)

hay x=-1

b) Ta có: \(x^3+x^2-x+2=0\)

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

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

\(\Leftrightarrow x+2=0\)

hay x=-2

a: \(x\left(x-1\right)+2x^2-2=0\)

=>\(x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x-1\right)\left(x+2x+2\right)=0\)

=>(x-1)(3x+2)=0

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

b: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

=>\(\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0\)

=>\(\left(3x+1\right)\left(3x-1-2x+3\right)=0\)

=>(3x+1)(x+2)=0

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

a: 

=>

=>

=>(x-1)(3x+2)=0

=>⎡⎣x=−13x=−2