a.\(\left(x+2\right)^2+x\left(2x-5\right)\)

=\(x^2+4x+4+2x^2-5x\)

=\(3x^2-x+4\)

b. \(\dfrac{x}{x+3}+\dfrac{2}{x-3}+\dfrac{-3x^2-9}{x^2-9}\)

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

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

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

a. 5x(x-3)(x+3)

b. (x-3)(x-4)

a. \(5x\left(x^2-9\right)\)

= \(5x\left(x+3\right)\left(x-3\right)\)

b.\(x^2-4x-3x+12\)

= \(\left(x^2-4x\right)-\left(3x-12\right)\)

\(=x\left(x-4\right)-3\left(x-4\right)\) 

= \(\left(x-3\right)\left(x-4\right)\)

Chiếc xe đạp sau khi được giảm giá khai trường còn:

12 000 000 x (100% - 10%)= 10 800 000 (đồng)

Anh Nam cần trả số tiền để mua chiếc xe đạp đó là:

10 800 000 x (100% - 8%) = 9 936 000 (đồng)

\(a,\\ \left(x+2\right)^2-x.\left(x-1\right)=10\\ \Leftrightarrow x^2+4x+4-x^2+x=10\\ \Leftrightarrow\left(x^2-x^2\right)+4x+x=10-4\\ \Leftrightarrow5x=6\\ \Leftrightarrow x=\dfrac{6}{5}\\ b,\\ x^3-6x^2+9x=0\\ \Leftrightarrow x.\left(x^2-6x+9\right)=0\\ \Leftrightarrow x.\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(5x^2-10x=5x.\left(x-2\right)\\ x^2-y^2-2x+2y=\left(x^2-y^2\right)-\left(2x-2y\right)\\ =\left(x-y\right).\left(x+y\right)-2.\left(x-y\right)\\ =\left(x+y-2\right).\left(x-y\right)\\ x^2+10x-y^2+25=\left(x^2+10x+25\right)-y^2\\ =\left(x+5\right)^2-y^2=\left(x+5-y\right).\left(x+5+y\right)\)

\(a,\\ \left(6x-7\right).\left(7x-1\right)=6x.7x-7x.7-6x.1-7.\left(-1\right)\\ =42x^2-49x-6x+7=42x^2-55x+7\\ b,\\ \left(4x-1\right)^2+\left(2x-5\right).\left(2x+5\right)=16x^2-8x+1+4x^2-25\\ =20x^2-8x-24\)

\(c,\\ \dfrac{x+5}{x}+\dfrac{x}{x-5}+\dfrac{25}{x^2-5x}\\ =\dfrac{\left(x-5\right).\left(x+5\right)}{x.\left(x-5\right)}+\dfrac{x.x}{x.\left(x-5\right)}+\dfrac{25}{x.\left(x-5\right)}\\ =\dfrac{x^2-25+x^2+25}{x.\left(x-5\right)}=\dfrac{2x^2}{x.\left(x-5\right)}=\dfrac{2x}{\left(x-5\right)}\left(ĐK:x\ne0;x\ne5\right)\)

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