1/2 + 1/2 = 2/3 + 1/3 = 3/4 + 1/4 = 4/5 + 1/5 = 2 - 1 = 1

= (sin^2x + cos^2x)^2 - 3sin^4x.cos^2x - 3sin^2x.cos^4x
= 1 - 3/4sin^2 (2x).sin^2x - 3/4sin^2(2x).cos^2x
= 1 - 3/4sin^2(2x)

c, x^3 - y^3 = xy + 8

1) Nếu x-y <= -1
(x -y)(x^2 + xy + y^2) = xy +8
=> (x -y)(x^2 + xy + y^2) <= -(x^2 + xy +y^2)
=> xy +8 <= -(x^2 + xy +y^2)
=> (x+y)^2 + 8 <=0 => Vô nghiệm

2) Nếu x-y =0 => x=y , Vô nghiệm

3) x- y>=1
=> (x -y)(x^2 + xy + y^2) >= x^2 + xy + y^2
=> xy + 8 >= x^2 + xy + y^2
=> x^2 + y^2 <=8
=> x^2 <=8

=> x=0 => y= -2
=> x= 1 => y + y^3 + 7 =0 (loại)

a, DKXD:\(x\ne2,x\ne-2\)

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

\(\Rightarrow x^3-x^2-2x-2x+4=x^3-x^2-4x+4\)\(=x^2\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\)

Để C=0 \(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

Khi |a|=5

a=5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(5-1\right)^2}{5-1}=4\)

a=-5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(-5-1\right)^2}{-5-1}=-6\)

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

để \(\dfrac{x-2}{x+2}=2\Leftrightarrow x-2=2\left(x+2\right)\Leftrightarrow x-2=2x+4\)\(\Leftrightarrow x-2x=4+2\Leftrightarrow-x=6\Leftrightarrow x=-6\)

Tại x=|3|

x=3\(\Rightarrow\dfrac{x-2}{x+2}=\dfrac{3-2}{3+2}=\dfrac{1}{5}\)

x=-3\(\Rightarrow\dfrac{x-2}{x+2}=\dfrac{-3-2}{-3+2}=5\)

a, ĐKXĐ:\(x\ne2,x\ne-2\)

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

a,\(x^2+2y^2+z^2-2xy-2y+2z+2=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2+2x+1\right)=0\)\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=-1\end{matrix}\right.\)