Toán Lớp 7: 1,Tính
a, B= $x^{3}$ -30x²-31x+1 tại x=31
b, C= $x^{5}$ – 15 $x^{4}$ +16x³-29x²+13x tại x=14
c, D= $x^{14}$-10 $x^{13}$+10 $x^{12}$ -10$x^{11}$+….+10$x^{2}$ -10x+10 tại x=9
2, Tìm 3 số tự nhiên liên tiếp, biết rằng nếu cộng 3 tích của 2 trong 3 số ấy ta được 242
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1,\\
a,\\
B = {x^3} – 30{x^2} – 31x + 1\\
= \left( {{x^3} – 31{x^2}} \right) + \left( {{x^2} – 31x} \right) + 1\\
= {x^2}\left( {x – 31} \right) + x\left( {x – 31} \right) + 1\\
= \left( {x – 31} \right)\left( {{x^2} + x} \right) + 1\\
x = 31 \Rightarrow x – 31 = 0 \Rightarrow \left( {x – 31} \right)\left( {{x^2} + x} \right) = 0\\
\Rightarrow B = 0 + 1 = 1\\
b,\\
C = {x^5} – 15{x^4} + 16{x^3} – 29{x^2} + 13x\\
= \left( {{x^5} – 14{x^4}} \right) + \left( { – {x^4} + 14{x^3}} \right) + \left( {2{x^3} – 28{x^2}} \right) + \left( { – {x^2} + 14x} \right) – x\\
= {x^4}.\left( {x – 14} \right) – {x^3}\left( {x – 14} \right) + 2{x^2}\left( {x – 14} \right) – x\left( {x – 14} \right) – x\\
= \left( {x – 14} \right).\left( {{x^4} – {x^3} + 2{x^2} – x} \right) – x\\
x = 14 \Rightarrow x – 14 = 0 \Rightarrow \left( {x – 14} \right).\left( {{x^4} – {x^3} + 2{x^2} – x} \right) = 0\\
\Rightarrow C = – x = – 14\\
c,\\
D = {x^{14}} – 10{x^{13}} + 10{x^{12}} – 10{x^{11}} + …. + 10{x^2} – 10x + 10\\
= \left( {{x^{14}} – 9{x^{13}}} \right) – \left( {{x^{13}} – 9{x^{12}}} \right) + \left( {{x^{12}} – 9{x^{11}}} \right) – \left( {{x^{11}} – 9{x^{10}}} \right) + …… + \left( {{x^2} – 9x} \right) – \left( {x – 9} \right) + 1\\
= {x^{13}}\left( {x – 9} \right) – {x^{12}}\left( {x – 9} \right) + {x^{11}}\left( {x – 9} \right) – {x^{10}}\left( {x – 9} \right) + …… + x\left( {x – 9} \right) – \left( {x – 9} \right) + 1\\
= \left( {x – 9} \right)\left( {{x^{13}} – {x^{12}} + {x^{11}} – {x^{10}} + …. + x – 1} \right) + 1\\
x = 9 \Rightarrow x – 9 = 0\\
\Rightarrow \left( {x – 9} \right)\left( {{x^{13}} – {x^{12}} + {x^{11}} – {x^{10}} + …. + x – 1} \right) = 0\\
\Rightarrow D = 1\\
2,
\end{array}\)
\left( {x – 1} \right).x + x\left( {x + 1} \right) + \left( {x + 1} \right).\left( {x – 1} \right) = 242\\
\Leftrightarrow \left( {{x^2} – x} \right) + \left( {{x^2} + x} \right) + \left( {{x^2} – x + x – 1} \right) = 242\\
\Leftrightarrow {x^2} – x + {x^2} + x + {x^2} – 1 = 242\\
\Leftrightarrow 3{x^2} – 1 = 242\\
\Leftrightarrow 3{x^2} = 243\\
\Leftrightarrow {x^2} = 81\\
\Leftrightarrow x = 9\,\,\,\,\,\left( {x \in {N^*}} \right)
\end{array}\)