\(2x^2y+4xy^2+2y^3-8\) 

\(=2y\left(x^2+2xy+y^2\right)-8\)

\(=2y\left(x+y\right)^2-8\) 

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

thank you huỳnh thanh phong

Hoàn thành bộ với một từ hoặc phara từ hộp???
 

Chưa có đề bài ạ

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

\(=\left(x^2+2\cdot5\cdot x+5^2\right)-\left(x^2+3x-2x-6\right)\)

\(=\left(x^2+10x+25\right)-\left(x^2+x-6\right)\)

\(=x^2+10x+25-x^2-x+6\)

\(=9x+31\)

\((x+5)^2-(x+3)(x-2)\\=(x^2+2\cdot x\cdot5+5^2)-[x(x-2)+3(x-2)]\\=(x^2+10x+25)-(x^2-2x+3x-6)\\=x^2+10x+25-(x^2+x-6)\\=x^2+10x+25-x^2-x+6\\=(x^2-x^2)+(10x-x)+(25+6)\\=9x+31\)

ko hiểu chi luôn á trời

1.C

2.B

3.B

4.B

5.B

\(-4x^2-24xy-36y^2\)

\(=-\left(4x^2+24xy+36y^2\right)\)

\(=-\left[\left(2x\right)^2+24xy+\left(6y\right)^2\right]\)

\(=-\left[\left(2x\right)^2+2\cdot2x\cdot6y+\left(6y\right)^2\right]\)

\(=-\left(2x+6y\right)^2\)

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

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

Chịu nhá

\(A=\dfrac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\dfrac{a^2b+5ab^2}{a^2-3ab}\\ =\dfrac{b\left[2a\left(a+5b\right)+\left(a+5b\right)\right]}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =\dfrac{b\left(a+5b\right)\left(2a+1\right)}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =2a+1\)

Do : \(a\in Z=>2a+1\)  là số nguyên lẻ (DPCM)

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\3-x\ne0\\x^2-9=\left(x-3\right)\left(x+3\right)\ne0\\x+2\ne0\end{matrix}\right.\\ < =>x\ne\left\{\pm3;-2\right\}\)

\(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\\ =\left[\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right].\dfrac{x-3}{x+2}\\ =\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)-\left(3-10x\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+2}\\ =\dfrac{2x^2-x-6x+3+x^2+3x-3+10x}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x^2+6x}{\left(x+3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x}{x+3}\)

\(\left(b\right):x^2-4=0< =>x^2=4\\ < =>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

Với \(x=2=>P=\dfrac{3.2}{2+3}=\dfrac{6}{5}\)

\(\left(a\right):A=\dfrac{x-5}{x-4}\left(x\ne4\right)\\ x^2-3x=0< =>x\left(x-3\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(TMDK\right)\)

Với \(x=0=>A=\dfrac{0-5}{0-4}=\dfrac{-5}{-4}=\dfrac{5}{4}\)

Với \(x=3=>A=\dfrac{3-5}{3-4}=\dfrac{-2}{-1}=2\)

\(\left(b\right):\dfrac{x+5}{2x}-\dfrac{x-6}{5-x}-\dfrac{2x^2-2x-50}{2x^2-10x}\left(x\ne\left\{0;5\right\}\right)\)

\(=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x\left(x-5\right)}\\ =\dfrac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-\left(2x^2-2x-50\right)}{2x\left(x-5\right)}\\ =\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\\ =\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x-5\right)}\\ =\dfrac{x-5}{2x}\)

\(\left(a\right):A=\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9}\left(x\ne\left\{\pm3\right\}\right)\\ =\dfrac{5}{x+3}+\dfrac{2}{x-3}-\dfrac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{5\left(x-3\right)+2\left(x+3\right)-\left(3x^2-2x-9\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

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

\(\left(b\right):\left|x-2\right|=1\\ =>\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)

Với \(x=1=>A=-\dfrac{3.1}{1+3}=-\dfrac{3}{4}\)

\(\left(b\right):A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right).\dfrac{x+2}{2}\\ =\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right].\dfrac{x+2}{2}\\ =\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{2}\\ =\dfrac{x-2x-4+x-2}{2\left(x-2\right)}\\ =\dfrac{-6}{2\left(x-2\right)}=\dfrac{3}{2-x}\)

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x^2-4\ne0\\2-x\ne0\\x+2\ne0\end{matrix}\right.< =>x\ne\pm2\)