a: \(\left(x-1\right)^3+27\)

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

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

b: \(\left(x-2\right)^3-8\)

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

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

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

3\(x^2\) + 12\(x\) - 2\(x\) - 8  - (\(x+1\))(\(x-2\)) -  [-(\(x-2\))](1- \(x\)) = 0 

                    3\(x^2\) + 10\(x\) - 8  -  (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0

                    3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\)  = 0

                    3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\)     = 0

                      \(x^2\) + 14\(x\) -  8 = 0

                    \(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0 

                     \(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57

                       (\(x+7\))(\(x\) + 7) =57

                       (\(x+7\))² = 57

                        \(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}

chưa học

Đề là giải pt hả bạn?

   \( \left(x+2\right)2-3x-8=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow2x+4-3x-8=1-x^2\)
\(\Leftrightarrow x^2-x-5=0 \) 
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{21}{4} =0\)
\(\Leftrightarrow\left(x-\frac{1+\sqrt{21}}{2}\right)\left(x-\frac{1-\sqrt{21}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{21}}{2}\\x=\frac{1-\sqrt{21}}{2}\end{cases}}\)
 

xy+4x+2y=-5

x(4+y)+2(y+4)=3

(4+y)(x+2)=3

=>4+y;x+2 thuộc Ư(3)={-1;1;3;-3}

còn lại lập bản rồi thử từng TH nhé

k đi

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

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

..............................................................................

..............<Giải thích như câu đầu>......................

.............................................................................

\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

......................................................................

...............<Giải thích như câu đầu>..............

.......................................................................

\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

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

..............................................................................

..............<Giải thích như câu đầu>......................

.............................................................................

\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

......................................................................

...............<Giải thích như câu đầu>..............

.......................................................................

\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

\(1,x^2-2x+5+y^2-4y=0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

\(2,y^2+2y+5-12x+9x^2=0\)

\(\Rightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Rightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(y+1\right)^2=0\\\left(3x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}y=-1\\x=\frac{2}{3}\end{cases}}}\)

\(3,x^2+20+9y^2+8x-12y=0\)

\(\Rightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+4\right)^2=0\\\left(3y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=\frac{2}{3}\end{cases}}}\)

Tui là Huỳnh Quốc Hữu nè, cái bài đó giải như sau:

x=-24; y=-10; z=-120.

Cách giải thì như trong vở nhá!

Hình như bn viết sai đề,là 1/x.(x+1) chứ

ukm mik xin lỗi mik viết sai đề đó

Đề sai nhé phải là x(x+1)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\Leftrightarrow\frac{x}{x+1}=\frac{2015}{2016}\Rightarrow x=2015\)

Vậy \(x=2015\)