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

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

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

b: Để A chia hết cho 4 thì x^2/x-3 chia hết cho 4 và x^2 chia hết cho x-3

=>x^2/x-3=4k(k thuộc Z) và x^2-9+9 chia hết cho x-3

=>\(x-3\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(x\in\left\{4;2;6;0;12;-6\right\}\)

Kết hợp ĐK chia hết cho 4, ta được: \(x\in\left\{4;12;-6\right\}\)

CÂU 1: 

\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

CÂU 2: 

\(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)

CÂU 3: 

\(\dfrac{15x\left(x+5\right)^3}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)^2}{4x}\)

CÂU 4: 

\(\dfrac{3xy+x}{9y+3}=\dfrac{x\left(3y+1\right)}{3\left(3y+1\right)}=\dfrac{x}{3}\)

CÂU 5: 

\(\dfrac{3xy+3x}{9y+9}=\dfrac{3x\left(y+1\right)}{9\left(y+1\right)}=\dfrac{x}{3}\)

CÂU 6: 

\(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)

CÂU 7:

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

CÂU 8: 

\(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

CÂU 9: 

\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\dfrac{2y}{3\left(x+y\right)^2}\)

a, Vì ABCD là hbh nên AB//CD

Do đó \(\widehat{A}+\widehat{D}=180^0\Rightarrow3\widehat{D}=180^0\Rightarrow\widehat{D}=60^0\Rightarrow\widehat{A}=120^0\)

Mà ABCD là hbh nên \(\left\{{}\begin{matrix}\widehat{A}=\widehat{C}=120^0\\\widehat{D}=\widehat{B}=60^0\end{matrix}\right.\)

b, Vì CE=CB nên tam giác CEB cân tại C

Do đó \(\widehat{B}=\widehat{CEB}\)

\(\Rightarrow\widehat{D}=\widehat{CEB}\left(1\right)\)

Mà ABCD là hbh nên AB//CD hay AE//CD

Do đó AECD là hình thang

Kết hợp (1) ta được AECD là hthang cân

a: BC=căn 6^2+8^2=10cm

bD là phân giác

=>AD/AB=CD/BC

=>AD/3=CD/5=(AD+CD)/(3+5)=8/8=1

=>AD=3cm; CD=5cm

b: Xét ΔBHA vuông tại H và ΔBAC vuông tại A có

góc B chung

=>ΔBHA đồng dạng với ΔBAC

=>BH/BA=BA/BC

=>BH*BC=BA^2

c: Xét ΔBHA có BI là phân giác

nên IH/IA=BH/BA

=>IH/IA=BA/BC=AD/DC

a) Ta có: \(\left(2x-3\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};-\dfrac{4}{3}\right\}\)

b) Ta có: \(x^3-3x^2+3x-1=\left(x-1\right)\left(x+1\right)\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)

Vậy: S={0;1;3}

c) Ta có: \(x^2+x=2x+2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy: S={-1;2}

d) Ta có: \(\left(x-1\right)^2=2\left(x^2-1\right)\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)Vậy: S={1;-3}

e) Ta có: \(2\left(x+2\right)^2-x^3-8=0\)

\(\Leftrightarrow2\left(x+2\right)^2-\left(x^3+8\right)=0\)

\(\Leftrightarrow2\left(x+2\right)\cdot\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)

Vậy: S={0;-2;4}

b4 

1, (x-y)³

3, [(x-y)(x+y)] - 4(x-y) 

= (x-y) [(x+y) - 4]

Bài 4: 

d: Ta có: \(x^2-y^2-2x-2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

e: Ta có: \(x^3-y^3-3x+3y\)

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

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

Đăng tách ra đi bạn

Bài 5: 

Theo đề, ta có:

\(\left(2x+5\right)^2-4x^2-12x=41\)

\(\Leftrightarrow20x-12x=41+25=66\)

hay \(x=8.25\left(m\right)\)

Chu vi là:

\(\left[\left(2\cdot8.25+5\right)^2+\left(4\cdot8.25^2+12\cdot8.25\right)\right]\cdot2=1667\left(m\right)\)