Giải:

a) \(\left(x-5\right)^2-16\)

\(=\left(x-5-4\right)\left(x-5+4\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

b) \(25-\left(3-x\right)^2\)

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

\(=\left(2+x\right)\left(8-x\right)\)

c) \(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)

\(=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

d) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

e) \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

Vậy ...

\(\left(x-5\right)^2-16\)

\(=\left(x-5\right)^2-4^2\)

\(=\left(x-5-4\right)\left(x-5+4\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

\(25-\left(3-x\right)^2\)

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

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

\(=\left(8-x\right)\left(2+x\right)\)

\(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=7^2\left(y-4\right)^2-3^2\left(y+2\right)^2\)

\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

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

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

\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)

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

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

c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

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

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

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

a, mình nghĩ đề là cm đẳng thức nhé 

\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)

Vậy ta có đpcm 

b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)

\(=-5y-9+xy=VP\)

Vậy ta có đpcm 

c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)

Vậy ta có đpcm 

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

\(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\)

\(P=\left(\frac{2x^2+y-2}{2y-x}.\frac{x+1}{2x^2+y-2}\right).\frac{1}{x+1}\)

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

thanks 

a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)

\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)