\(ĐKXĐ:x\ne\frac{3}{2}\)

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+2x+7=x^2+10\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\left(KTMĐKXĐ\right)\)

Vậy phương trình vô nghiệm

ĐKXĐ: x khác 3/2

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)

<=> \(\frac{x^2+4x+4}{2x-3}-1=\frac{x^2+10}{2x-3}\)

<=> x^2 + 4x + 4 - 2x + 3 = x^2 + 10

<=> x^2 + 4x + 4 - 2x + 3 - x^2 - 10 = 0

<=> 2x - 3 = 0

<=> 2x = 0 + 3

<=> 2x = 3

<=> x = 3 (ktmdk)

=> pt no

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

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

\(=>\frac{2}{x+1}=\frac{1}{x+1}\)

\(=>2x+2=x+1\)

\(=>2x-x=1-2=-1\)

\(=>x=-1\)

vậy nghiệm của phương trình trên là {-1}

À quên ĐKXĐ của câu a là \(x\ne-1\)

Nên \(x\in\varnothing\)nhé :v

\(\left(2x^2+1\right)\left(4x-2\right)=\left(2x^2+1\right)\left(x-12\right)\)

\(\Leftrightarrow8x^3-6x^2+4x-3=2x^3-24x^2+x-12\)

\(\Leftrightarrow8x^3-6x^2+4x-3-2x^3+24x^4-x+12=0\)

\(\Leftrightarrow6x^3+18x^2+3x+9=0\)

\(\Leftrightarrow3\left(x+3\right)\left(2x^2+1\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

Bạn chắc bạn viết đúng đề bài không?

\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)

Quy đồng mẫu chung :

\(\frac{2.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(4x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

Sau đó ta khử mẫu:

\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)

\(\Rightarrow\)\(2x^2+2x+2x^2+x-4x^2=-1-2+3\)

\(\Rightarrow\)\(3x=0\)

\(\Rightarrow\)\(x=0\)

Vậy bạn tự kết luận 

ĐKXĐ:    \(x\ne1\)

           \(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)

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

\(\Leftrightarrow\)\(\frac{2x^2+2x+2}{x^3-1}+\frac{2x^2+x-3}{x^3-1}=\frac{4x^2-1}{x^3-1}\)

\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)

\(\Leftrightarrow\)\(4x^2+3x-1=4x^2-1\)

\(\Leftrightarrow\)\(3x=0\)

\(\Leftrightarrow\)\(x=0\)   (thỏa mãn)

Vậy....

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

\(\Leftrightarrow2x^2+2x+2+2x^2-2x+3x-3\)\(=4x^2-1\)

\(\Leftrightarrow2x^2+2x^2-4x^2+2x-2x+3x=-1-2+3\)

\(\Leftrightarrow3x=0\)

\(\Leftrightarrow x=0\left(nhận\right)\)

Vậy S= {0}

a)\(\frac{x+3}{6}\)+\(\frac{x-2}{10}\)>\(\frac{x+1}{5}\)

<=> \(\frac{5\left(x+3\right)}{30}\)+\(\frac{3\left(x-2\right)}{30}\)>\(\frac{6\left(x+1\right)}{30}\)

<=>5(x+3)+3(x-2)>6(x+1)

<=>5x+15+3x-6>6x+6

<=>8x-6x           >6-15+6

 <=>2x               >-3

<=>x                  >-1,5    

Vậy tập nghiệm của bất phương trình là {x/x>-1,5}

b)(x+1)(2x-2)-3<-5x-(2x+1)(3-x)

<=> 2x\(^2\)-2x+2x-2-3<-5x-6x+2x\(^2\)-3+x

<=>2x\(^2\)-2x\(^2\)+5x+6x-x<2+3-3

<=>10x <2

<=>x   <\(\frac{1}{5}\) 

Vậy tập nghiệm của bất phương trình là {x/x<\(\frac{1}{5}\)}

thank nhiều, iu <3

a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)

\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)

\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)

\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)

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

\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2.2x}{2\left(x-3\right)\left(x+1\right)}=0\)

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

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

\(\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)

=>\(2x^2-6x=0\)

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

=>\(x=0\)

\(x=3\)