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ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)

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

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)

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

\(\dfrac{x^2-7x+12}{x^2-6x+8}+\dfrac{x^2-4x+2}{x^2-6x+8}=\dfrac{-x^2+6x-8}{x^2-6x+8}\)

\(\dfrac{x^2-7x+12+x^2-4x+2+x^2-6x+8}{x^2-6x+8}=0\)

\(\dfrac{3x^2-17x+22}{x^2-6x+8}=0\)

\(\dfrac{\left(3x-11\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

\(\dfrac{3x-11}{x-4}=0\)

Với x khác 4 

=> 3x -11 = 0 

=> \(x=\dfrac{11}{3}\)

a,

(x²-x+1)(x+1)-x³+3x=15

x³-x²+x+x²-x+1-x³+3x=15

x³-x³-x²+x²+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

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

x²-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x²-5)(x+2)+5x=2x²+17

x³+2x²-5x-10+5x=2x²+17

x³+2x²-5x+5x-10=2x²+17

x³+2x²-10=2x²+17

x³-10=17

x³=17+10

x³=27

\(\Rightarrow x=3\)(Vì : 3³=27)

_k_ nhé bn

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)

\(\Leftrightarrow x^2+4x-6=4x+3\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)

Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html

\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

Lời giải:
$x^2+x-6=(x^2-2x)+(3x-6)=x(x-2)+3(x-2)=(x-2)(x+3)$

`A=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2)`

`=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x`

`=(x^4-x^4)+(x^3-x^3)+(3x^2-x^2-2x^2)+(2x-2x)+2`

`=2`

\(x^2\) + (\(x\) + 2)(11\(x\) - 7) = 4

        (\(x\) + 2)(11\(x\) - 7) + \(x^2\) - 4 = 0

        (\(x+2\))(11\(x-7\)) + (\(x-2\))(\(x+2\)) = 0

        (\(x+2\))(11\(x\) - 7 + \(x-2\)) = 0

        (\(x\) + 2)(12\(x\) - 9) = 0

         \(\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=-2\\x=\dfrac{9}{12}\end{matrix}\right.\)

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

.

\(\dfrac{15}{7}:\dfrac{7}{9}-\dfrac{4}{7}\times\dfrac{2}{9}-\dfrac{4}{7}\times\dfrac{4}{9}\)

\(=\dfrac{15}{7}\times\dfrac{9}{7}-\dfrac{4}{7}\times\dfrac{2}{9}-\dfrac{4}{7}\times\dfrac{4}{9}\)

\(=\dfrac{135}{49}-\dfrac{8}{63}-\dfrac{16}{63}\)

\(=\dfrac{1215}{441}-\dfrac{56}{441}-\dfrac{112}{441}\)

\(=\dfrac{1215-56-112}{441}\)

\(=\dfrac{1047}{441}=\dfrac{349}{147}\)

15/7 : 7/9 - 4/7 x 2/9 - 4/7 x 4/9

= 135/49 - 8/63 - 16/63

= 1159/441 - 16/63

= 1047/441 = 349/147