Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)

\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)

\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)

\(\Leftrightarrow2x^2+5=0\)(vô lý)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)

\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)

Vậy pt vô nghiệm

\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)

=13/54

Các bạn cố gắng giúp mình nha . Mình xin chân thành cảm ơn 

\(\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}\)

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

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

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

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

b) Bạn có thể viết kiểu latex được không ạ ?

a) => 4/3x = 7/9 - 4/9 = 1/3

=> x = 1/3 : 4/3 = 1/4

b) => 5/2 - x = 9/14 : (-4/7) = -9/8

=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8

c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12

=> x = 23/12 : 3 = 23/36

D) => -5/6 - x = 1/4

=> x = -5/6 - 1/4 = -13/12

a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)

\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{4}{3}\)

\(x=\dfrac{1}{4}\)

b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)

\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)

\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)

\(x=\dfrac{29}{8}\)

c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)

\(x=\dfrac{23}{12}:3\)

\(x=\dfrac{23}{36}\)

d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(x=-\dfrac{13}{12}\)

ta có:\(x^3+x^2+2x^2+2x+2x+2=0\)0

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

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

Do \(x^2+2x+2\ne0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

vậy phương trình trên có tập nghiệm là :S=(-1)