Toán Lớp 8: rút gọn: (x ² – 5x + 1) ² + 2(5x-1)(x ² – 5x + 1) + (5x-1) ²
CMR: 1000² + 1003 ² + 1005 ² + 1006 ²= 1001 ² + 1002 ² + 1004 ² + 1007 ²
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Toán Lớp 8: rút gọn: (x ² – 5x + 1) ² + 2(5x-1)(x ² – 5x + 1) + (5x-1) ²
CMR: 1000² + 1003 ² + 1005 ² + 1006 ²= 1001 ² + 1002 ² + 1004 ² + 1007 ²
Comments ( 1 )
a){\left( {{x^2} – 5x + 1} \right)^2} + 2.\left( {5x – 1} \right)\left( {{x^2} – 5x + 1} \right) + {\left( {5x – 1} \right)^2}\\
= {\left( {{x^2} – 5x + 1 + 5x – 1} \right)^2}\\
= {\left( {{x^2}} \right)^2}\\
= {x^4}\\
b)Giả\,sử:\\
{1000^2} + {1003^2} + {1005^2} + {1006^2}\\
= {1001^2} + {1002^2} + {1004^2} + {1007^2}\\
\Leftrightarrow {1001^2} – {1000^2} + {1007^2} – {1006^2}\\
= {1003^2} – {1002^2} + {1005^2} – {1004^2}\\
\Leftrightarrow \left( {1001 – 1000} \right)\left( {1001 + 1000} \right) + \left( {1007 – 1006} \right)\left( {1007 + 1006} \right)\\
= \left( {1003 – 1002} \right)\left( {1003 + 1002} \right) + \left( {1005 – 1004} \right)\left( {1005 + 1004} \right)\\
\Leftrightarrow 2001 + 2013 = 2005 + 2009\\
\Leftrightarrow 4014 = 4014\left( {tmdk} \right)\\
Vay\,{1000^2} + {1003^2} + {1005^2} + {1006^2}\\
= {1001^2} + {1002^2} + {1004^2} + {1007^2}
\end{array}$