c:

Đặt \(x^2+3x=t\)

\(\left(t+1\right)\left(t-3\right)-5=t^2-2t-8=\left(t-1\right)^2-9=\left(t-4\right)\left(t+2\right)\)

\(\Rightarrow\left(x^2+3x-4\right)\left(x^2+3x+2\right)=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

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

đk x khác 2 ;3 

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

\(\Rightarrow\left(x+3\right)\left(x-3\right)+\left(x+2\right)\left(x-2\right)=2\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow x^2-9+x^2-4=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow2x^2-13=2x^2-10x+12\Leftrightarrow10x-25=0\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

\(\dfrac{x+3}{x-2}+\dfrac{x+2}{x-3}=2\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2\left(x-3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ \Leftrightarrow\left(x^2-9\right)+\left(x^2-4\right)=2\left(x^2-2x-3x+6\right)\\ \Leftrightarrow2x^2-13=2x^2-10x+12\\ \Leftrightarrow-10x+12=-13\\ \Leftrightarrow-10x=-25\\ \Leftrightarrow x=\dfrac{25}{10}\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

a) \(A=5x^{2n}y^{6-n}\)

\(B=-\dfrac{1}{3}x^{n+3}y^n\)

\(\Rightarrow\dfrac{A}{B}=-15x^{n-3}y^{6-2n}\)

Để \(A⋮B\) khi

\(\Rightarrow\left\{{}\begin{matrix}n-3\in N\\6-2n\in N\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n\in\left\{3;4;5;...\right\}\\n\in\left\{0;1;2;3\right\}\end{matrix}\right.\)

\(\Rightarrow n=3\)

b) \(C=7x^9y^{n+2}-5x^{n+1}y^n\)

\(D=\dfrac{1}{2}x^ny^9\)

\(\Rightarrow\dfrac{C}{D}=14x^{9-n}y^{n-7}-10xy^{n-9}=2xy^{n-9}\left(7x^{8-n}y^2-5\right)\)

Để \(C⋮D\) khi

\(\Rightarrow\left\{{}\begin{matrix}n-9\in N\\8-n\in N\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n\in\left\{9;10;11;...\right\}\\n\in\left\{0;1;2;3;4;5;6;7;8\right\}\end{matrix}\right.\)

\(\Rightarrow n\in\varnothing\)

a; \(x^2\) - 6\(x\) + 8 

= \(x^2\) - 2\(x\)  - 4\(x\) + 8

= (\(x^2\) - 2\(x\)) - (4\(x\) - 8)

= \(x\)(\(x\) - 2) - 4(\(x\) - 2)

= (\(x-2\))(\(x\) - 4)

4\(x^2\) + 4\(x\) - 3

= 4\(x^2\) - 2\(x\) + 6\(x\) - 3

= (4\(x^2\) - 2\(x\)) + (6\(x\) - 3)

= 2\(x\)(2\(x\) - 1) + 3(2\(x\) - 1)

= (2\(x\) - 1)(2\(x\) + 3)

Dòng cuối \(AB\perp CB\) nha mình sửa ko đc:")

Xét \(\Delta OCD\) có \(\widehat{COD}+\widehat{OCD}+\widehat{ODC}=180^o\Rightarrow\widehat{OCD}+\widehat{ODC}=180^o-\widehat{COD}=180^o-110^o=70^o\)

CO, DO lần lượt là phân giác của \(\widehat{BCD},\widehat{ADC}\) nên \(\widehat{BCD}=2\widehat{OCD},\widehat{ADC}=2\widehat{ODC}\Rightarrow\widehat{BCD}+\widehat{ADC}=2\left(\widehat{OCD}+\widehat{ODC}\right)=2.70^o=140^o\)

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\Rightarrow\widehat{A}+\widehat{B}=360^o-\left(\widehat{C}+\widehat{D}\right)=360^o-140^o=220^o\Rightarrow\widehat{A}=220^o-\widehat{B}\)

\(\widehat{A}-\widehat{B}=40^o\Rightarrow\left(220^o-\widehat{B}\right)-\widehat{B}=40^o\Rightarrow220^o-2\widehat{B}=40^o\Rightarrow2\widehat{B}=180^o\Rightarrow\widehat{B}=90^o\)

Do đó \(AB\perp CD\)

Bài 1:

Thay $3=x^2+xy+y^2$ vào PT(2) thì:

$2x^3=(x+y)(x^2+xy+y^2-2xy)$

$\Leftrightarrow 2x^3=(x+y)(x^2-xy+y^2)=x^3+y^3$

$\Leftrightarrow x^3=y^3\Leftrightarrow x=y$. 

Thay vào PT(1) thì: $3x^2=3\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

$\Rightarrow y=\pm 1$ (tương ứng)

Vậy HPT có nghiệm $(x,y)=(\pm 1, \pm 1)$

Bài 2:

Thay $2=xy(x+y)$ vào PT(2) thì:

$x^3+y^3+3xy(x+y)=8y^3$

$\Leftrightarrow (x+y)^3=(2y)^3$

$\Leftrightarrow x+y=2y\Leftrightarrow x=y$. 

Thay vào PT(1): $x^2.2x=2$

$\Leftrightarrow 2x^3=2\Leftrightarrow x^3=1\Leftrightarrow x=1$.

$\Rightarrow y=x=1$

Vậy HPT có nghiệm $(x,y)=(1,1)$