Rút gọn biểu thức sau:

(a) \[5{x^2}{y^4}.\left( {\frac{{ - 1}}{{20}}x{y^3}} \right)\]

(b) \[\left( {{x^3}y + 4{x^2}{y^2} - 5{x^2}{y^3}} \right):xy - x\left( {x - 4{y^2}} \right)\].

(c) \[{\left( {x + 1} \right)^2} - \left( {2x - 1} \right)\left( {x + 3} \right) + \left( {x - 5} \right)\left( {x + 5} \right)\].

Ta có \({x^2} + {y^2} + {z^2} + 2x - 2y - 4z + 6 = 0.\)

\({x^2} + 2x + 1 + {y^2} - 2y + 1 + {z^2} - 4z + 4 = 0\)

\({\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {z - 2} \right)^2} = 0\)

Khi đó: \(\left\{ \begin{array}{l}x + 1 = 0\\y - 1 = 0\\z - 2 = 0\end{array} \right.\) hay \(\left\{ \begin{array}{l}x = - 1\\y = 1\\z = 2\end{array} \right.\)

Thay \[x = - 1\,;\,\,y = 1\,;\,\,z = 2\] vào \[M\], ta được:

\(M = {\left( {x + 1} \right)^{2022}} + {y^{2023}} + {\left( {z - 1} \right)^{2024}}\)

\(M = {\left( { - 1 + 1} \right)^{2022}} + {1^{2023}} + {\left( {2 - 1} \right)^{2024}}\)

\(M = 1 + 1 = 2.\)

Vậy \(M = 2.\)

a) \[4x\left( {x - 3} \right) - 4{x^2} + 1 = 9\]

\[4{x^2} - 12x - 4{x^2} + 1 = 9\]

\[ - 12x = 8\]

\[x = \frac{{ - 2}}{3}\]

Vậy \[x = \frac{{ - 2}}{3}\]

b) \[\left( {2x - 5} \right)\left( {3x + 4} \right) - 6x\left( {x - 3} \right) = 5x\].

\[6{x^2} + 8x - 15x - 20 - 6{x^2} + 18x = 5x\]

\[11x - 20 = 5x\]

\[6x = 20\]

\[x = \frac{{10}}{3}\]

Vậy \[x = \frac{{10}}{3}\]

c) \[{\left( {x + 3} \right)^2} + {\left( {x - 7} \right)^2} = x\left( {2x - 5} \right)\].

\[{x^2} + 6x + 9 + {x^2} - 14x + 49 = 2{x^2} - 5x\]

\[3x = 58\]

\[x = \frac{{58}}{3}\]

Vậy\[x = \frac{{58}}{3}\].