Bài 1:

a: \(x\left(x+y\right)+5y-x^2\)

\(=x^2+xy+5y-x^2\)

=xy+5y

b: \(\left(x-2\right)\left(y+1\right)-xy+4\)

\(=xy+x-2y-2-xy+4\)

=-2y+x+2

c: \(\dfrac{\left(4x^2y+12xy^2-8xy\right)}{2xy}\)

\(=\dfrac{2xy\cdot2x+2xy\cdot6y-2xy\cdot4}{2xy}\)

=2x+6y-4

d: \(\left(x-4\right)^2+8x-7\)

\(=x^2-8x+16+8x-7\)

\(=x^2+9\)

a,\(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}=\dfrac{3-x+2x-8}{x-5}=\dfrac{x-5}{x-5}=1\)

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

ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)

a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

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

\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)

\(=\dfrac{2x}{3x-y}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

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

\(=\dfrac{x+5}{x.\left(x+5\right)}\)

\(=\dfrac{1}{x}\)

a) Ta có: \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

\(=\dfrac{x\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{x\left(3x+y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{2xy}{\left(3x+y\right)\left(3x-y\right)}\)

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

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

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

\(=\dfrac{2x}{3x-y}\)

b) Ta có: \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x\left(x+5\right)}-\dfrac{3x}{x\left(x+5\right)}\)

\(=\dfrac{4x+5-3x}{x\left(x+5\right)}\)

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

\(=\dfrac{1}{x}\)

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)

a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y

=5x^3-7x^2y+2xy^2+5x-2y

b: =(x^2-1)(x+2)

=x^3+2x^2-x-2

c: =1/2x^2y^2(4x^2-y^2)

=2x^4y^2-1/2x^2y^4

d: =(x^2-1/4)(4x-1)

=4x^3-x^2-x+1/4

e: =x^2-2x-35+(2x+1)(x-3)

=x^2-2x-35+2x^2-6x+x-3

=3x^2-7x-38

giúp mình nha mình cần gấp, cảm ơn mọi người trước

a) (2x+3)²

=4x^2+12x+9

b) (x-2/5)³

=x^3-1.2x^2+0.48x-0.064

c) (4x²+1)³

=(4x^2)^3+12x^4+12x^2+1

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

b) \(=\dfrac{1}{x+2}+\dfrac{3}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\)

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

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)