1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3) 
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8 
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5 
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27 
<=> - 3x^2 - 18x - 22 = 0 
<=> 3x^2 + 18x + 22 = 0 

Nửa chu vi mảnh đất là: 

                                               120 : 2 = 60 (m)

Chiều dài hơn chiều rộng là:

                                               5 + 5 = 10 (m)

Chiều rộng là:

                                          ( 60 - 10 ) : 2 = 25 (m)

Chiều dài là:

                                                25 + 10 = 35 (m)

Diện tích là:

                                               25  35 = 875 ( )

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

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

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

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

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

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

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

Giải:

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

     \(\dfrac{-5}{6}-x=\dfrac{1}{4}\)

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

               \(x=\dfrac{-13}{12}\) 

b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\) 

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\) 

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)  

             \(x-\dfrac{1}{3}=\dfrac{2}{3}:2\) 

             \(x-\dfrac{1}{3}=\dfrac{1}{3}\) 

                    \(x=\dfrac{1}{3}+\dfrac{1}{3}\) 

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

c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\) 

           \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\) 

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\) 

d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\) 

\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\) 

            \(x.\dfrac{5}{6}=\dfrac{29}{8}\) 

                \(x=\dfrac{29}{8}:\dfrac{5}{6}\) 

                \(x=\dfrac{87}{20}\)

Giúp

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(x=\dfrac{6}{20}=\dfrac{3}{10}\)

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

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

\(x=\dfrac{9}{6}=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(x=\dfrac{1}{7}+\dfrac{4}{35}+\dfrac{2}{5}\)

\(x=\dfrac{23}{35}\)

.

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(\Leftrightarrow x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

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

\(\Leftrightarrow x=\dfrac{5}{6}+\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(\Leftrightarrow x=\dfrac{1}{7}+\dfrac{2}{5}+\dfrac{4}{35}\)

\(\Leftrightarrow x=\dfrac{23}{35}\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

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

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

a) x=8/9 - 1/6
    x= 13/18
b) x=7/10+1/6
   x= 13/15
c)  5/2-x=3/4
   -x=3/4 - 5/2
-x=-7/4
  x=7/4

like và theo dõi mik đuy

ai biết, t lớp 8 mà , là lại bài lớp 4 hok biết đúng ko nữa =)

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

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