a: ĐKXĐ: \(x\le0\)

b: ĐKXĐ: \(x\le2\)

\(a,ĐK:-3x\ge0\Leftrightarrow x\le0\left(-3< 0\right)\\ b,ĐK:4-2x\ge0\Leftrightarrow-2x\ge-4\Leftrightarrow x\le2\\ c,ĐK:\dfrac{1}{2x-5}\ge0\Leftrightarrow2x-5>0\left(1>0;2x-5\ne0\right)\\ \Leftrightarrow x>\dfrac{5}{2}\\ d,ĐK:\dfrac{4x+7}{-3}\ge0\Leftrightarrow4x+7\le0\left(-3< 0\right)\Leftrightarrow x\le-\dfrac{7}{4}\)

\(8,=\left(2x-3\right)\left(2x+3\right)\\ 9,=\left(1-5a^2\right)\left(1+5a^2\right)\)

8) \(-9+4x^2=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

9) \(1-25a^4=1-\left(5a^2\right)^2=\left(1-5a^2\right)\left(1+5a^2\right)\)

ko bt

a: \(Q=\dfrac{2x^2-4x+x-3-6}{\left(x-3\right)\left(x-2\right)}\cdot\dfrac{x-2}{x^2+1}=\dfrac{2x^2-3x-9}{x-3}\cdot\dfrac{1}{x^2+1}\)

\(=\dfrac{2x^2-6x+3x-9}{x-3}\cdot\dfrac{1}{x^2+1}=\dfrac{2x+3}{x^2+1}\)

b: Để Q>0 thì 2x+3>0

hay x>-3/2

Bài 2:

a) Q=(\(\dfrac{2x}{x-3}-\dfrac{1}{2-x}-\dfrac{6}{x^2-5x+6}\)):\(\dfrac{x^2+1}{x-2}\)

       =(\(\dfrac{2x}{x-3}+\dfrac{1}{x-2}-\dfrac{6}{x^2-2x-3x+6}\)):\(\dfrac{x^2+1}{x-2}\)

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

        =\(\dfrac{2x\left(x-2\right)+x-3-6}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x^2-4x+x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x^2-3x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x^2-6x+3x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

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

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

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

         =\(\dfrac{2x+3}{x^2+1}\)

Ta có

\(\frac{1}{2}=\frac{1\times3\times5}{2\times3\times5}=\frac{15}{30}\)

\(\frac{1}{3}=\frac{1\times2\times5}{3\times2\times5}=\frac{10}{30}\)

\(\frac{2}{5}=\frac{2\times2\times3}{5\times2\times3}=\frac{12}{30}\)

Hok tốt !!!!!!!!!!!!!!!!!!!

a: \(=5x^2-10x-5x^2+7x=-3x\)

b: \(=2x^3+3xy^2-4y-3xy^2=2x^3-4y\)

giả sử số đó là 19

19-10=9đáp số 19

x-10=1 số nào đó

10+1=11

Đ/S:11

Ta có:

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

\(4-4x=-4\left(x-1\right)\)

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

\(\Rightarrow\) MTC là \(3.\left(-4\right).\left(x-1\right)\left(x+1\right)=-12\left(x-1\right)\left(x+1\right)\)

Do đó:

\(\dfrac{11x}{3x-3}=\dfrac{11x}{3\left(x-1\right)}=\dfrac{11x.\left(-4\right).\left(x+1\right)}{3\left(x-1\right).\left(-4\right)\left(x+1\right)}=\dfrac{-44x\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{5}{4-4x}=\dfrac{5}{-4\left(x-1\right)}=\dfrac{5.3\left(x+1\right)}{-4\left(x-1\right).3\left(x+1\right)}=\dfrac{15\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2x}{x^2-1}=\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x.\left(-12\right)}{-12\left(x-1\right)\left(x+1\right)}=\dfrac{-24x}{-12\left(x-1\right)\left(x+1\right)}\)

\(15x^2y^5-10x^3y^4=5x^2y^4\left(3y-2x\right)\)

\(4x\left(x-2y\right)+7\left(2y-x\right)=4x\left(x-2y\right)-7\left(x-2y\right)=\left(x-2y\right)\left(4x-7\right)\)

\(5x^3+20x^2y+20xy^2=5x\left(x^2+4xy+4y^2\right)=5x\left(x+2y\right)^2\)

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