Bài 1: Phân thức đại số.

\(a,VT=\dfrac{x^2y^3\cdot7xy}{5\cdot7xy}=\dfrac{7x^3y^4}{35xy}=VP\\ b,VT=\dfrac{x^2\left(x+2\right):x\left(x+2\right)}{x\left(x+2\right)^2:x\left(x+2\right)}=\dfrac{x}{x+2}=VP\\ c,VT=\dfrac{\left(3-x\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}=\dfrac{\left(x-3\right)^2}{9-x^2}=VP\\ d,VT=\dfrac{x\left(x^2-4\right)}{5\left(2-x\right)}=\dfrac{x\left(x-2\right)\left(x+2\right)}{-5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}=VP\\ e,VT=\dfrac{5y\cdot4x}{7\cdot4x}=\dfrac{20xy}{28x}=VP\\ f,VP=\dfrac{3x\cdot\left(x+5\right)}{2\cdot\left(x+5\right)}=VP\\ g,VP=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}=VT\)

\(h,VT=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2=\dfrac{\left(x-2\right)\left(x-1\right)}{x-1}=\dfrac{x^2-3x+2}{x-1}=VP\\ i,VT=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)

\(a,ĐK:5x-10\ne0\Leftrightarrow x\ne2\\ b,ĐK:-2x-6\ne0\Leftrightarrow x\ne-3\\ c,ĐK:-2x+22\ne0\Leftrightarrow x\ne11\)

\(a,A=\dfrac{\left(6x^2+3x\right)\left(2x-1\right)}{4x^2-1}=\dfrac{3x\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=3x\\ b,A=\dfrac{\left(4x^2-3x-7\right)\left(2x+3\right)}{4x-7}=\dfrac{\left(x+1\right)\left(4x-7\right)\left(2x+3\right)}{4x-7}=\left(x+1\right)\left(2x+3\right)\\ c,A=\dfrac{\left(4x^2-7x+3\right)\left(x^2+2x+1\right)}{x^2-1}=\dfrac{\left(x-1\right)\left(4x-3\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\left(4x-3\right)\left(x+1\right)\\ d,A=\dfrac{\left(2x^2-3x-2\right)\left(x^2+2x\right)}{x^2-2x}=\dfrac{x\left(x+2\right)\left(x-2\right)\left(2x+1\right)}{x\left(x-2\right)}=\left(x+2\right)\left(2x+1\right)\)

\(\dfrac{x^2+x-2}{x^2-1}=\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x+1}=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x^2-4}{x^2-x-2}\)

a) \(\dfrac{5y}{7}=\dfrac{5y.4x}{7.4x}=\dfrac{20xy}{28x}\)

b) \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)

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

d) \(\dfrac{x^2-x-2}{x+1}=\dfrac{x\left(x-2\right)+1\left(x-2\right)}{x+1}=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2=\dfrac{\left(x-2\right)\left(x-1\right)}{x-1}=\dfrac{x^2-3x+2}{x-1}\)

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

ý bạn là tìm x hay sao?

\(a,\Leftrightarrow\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=\dfrac{x^2-3x-2}{x-1}\left(x\ne\pm1\right)\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=x^2-3x-2\\ \Leftrightarrow x^2-3x+2=x^2-3x-2\\ \Leftrightarrow2=-2\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\\ \Leftrightarrow x+2=x+2\\ \Leftrightarrow x\in R\)

 alo cho tui hỏi bạn có phải Dương Ngọc Lan Hương Trường THCS Minh Thuận 3 k dọ

\(a,\Leftrightarrow A=\dfrac{8x^3y^4}{2x^3y^2}=4y^2\\ b,\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x^2-1\right)\left(x+2\right)}\\ \Leftrightarrow A=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x-1}\)