đề sai òi                                              

A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

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

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)

<=> 5(x - 4) + 2x = 2(x +1)

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

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)

phương trình tương đương với 1+\(\frac{1}{x}+1+\frac{1}{x+3}\)=1+\(\frac{1}{x+1}+1+\frac{1}{x+2}\)\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+3}=\frac{1}{x+2}+\frac{1}{x+1}\)

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

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

\(\Leftrightarrow\left(2x+3\right)\left(\frac{2}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)\(\Leftrightarrow2x+3=0\Leftrightarrow x=\frac{-3}{2}\)

\(\left(x-1\right)\left(x+1\right)-2\left(2x+3\right)\le\left(x-2\right)^2+x\)

\(\Leftrightarrow x^2-1-4x-6\le x^2-4x+4+x\)

\(\Leftrightarrow x^2-4x-7\le x^2-3x+4\)

\(\Leftrightarrow x^2-4x-x^2+3x\le7+4\)

\(\Leftrightarrow-x\le11\)

\(\Leftrightarrow x\le-11\)

biết đừng đăng anh à

Phân tích  : x²-3x +2=(x-1)(x-2) , x²-4x +3 = (x-1 )(x-3) ,  điều kiện  : x # 1, x # 2 ,x # 3

pt tương đương với  : \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{2x+5+x+1}{\left(x-1\right)\left(x-3\right)}\)

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

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

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

                             <=> x=0 hoặc x=1/2

\(1,\frac{7x-3}{x-1}=\frac{2}{3}\)    ĐKXĐ : \(x\ne1\)

\(\Leftrightarrow\frac{3\left(7x-3\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Rightarrow21x-2x=9-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)(TM)

kl :....

\(3,\frac{1}{x-2}+3=\frac{x-3}{2-x}\)   ĐKXĐ : \(x\ne2\)

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

\(\Leftrightarrow1+3x-6=3-x\)

\(\Leftrightarrow3x+x=-1+6-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=2\)(TM)

KL : ....

Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4