1) \(x^3-x^2=4x^2-8x+4\)

\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)

\(\Leftrightarrow x^2-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-2x.2+2^2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Câu 2 nha

\(a,x^4+2x^3+x^2\)

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

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

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

\(=\left(x+y\right)^3-\left(x+y\right)\)

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

Từ đầu bài 

=> 1.\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\) \(+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)

=> \(\left(a-b\right)\left(a+b\right)+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}+b^{64}\)

=> \(\left(a^2-b^2\right)\left(a^2+b^2\right)+...+\left(a^{32}+b^{32}\right)\)= a^64 + b^64

tương tự sẽ ra kết quả cuối là \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)

ta có x^2 + 2y^2 +z^2 -2xy -2y -4z +5 =0

=> (x^2 - 2xy +y^2) + (y^2 -2y +1) + (z^2 -4z +4) =0

=> (x-y)^2 + (y-1)^2 +(z-2)^2 =0

=> x=y , y=1 , z=2 

=> A= (1-1)^2018 + (1-1)^2019 + ( 2-1)^2020 => A= 1

nghĩ thế !

a) \(x-y=-3\)

\(\Rightarrow\left(x-y\right)^2=9\)

\(\Leftrightarrow x^2-2xy+y^2=9\)

\(\Leftrightarrow x^2+y^2-20=9\)

\(\Leftrightarrow x^2+y^2=29\)

b) \(\left(x-y\right)^3=-27\)

\(\Leftrightarrow x^3-y^3-3xy\left(x+y\right)=-17\)

\(\Leftrightarrow x^3-y^3-30\left(x+y\right)=-17\)

WHAT ??? 

   5x.(x-3)-2.(3-x)

= 5x.(x-3) + 2.(x-3)

= (x-3).(5x+2)