Question

Đã trả lời bởi chuyên gia

Answer 1

\(\begin{array}{l}
1)x\left( {x + 2} \right) + {x^2} = - 2{\rm{x}}\\
\Leftrightarrow x\left( {x + 2} \right) + {x^2} + 2{\rm{x}} = 0\\
\Leftrightarrow x\left( {x + 2} \right) + x\left( {x + 2} \right) = 0\\
\Leftrightarrow 2{\rm{x}}\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
vay:S = \{ 0; - 2\} \\
2){x^4} - 9{{\rm{x}}^3} = {x^2} - 9{\rm{x}}\\
\Leftrightarrow {x^3}\left( {x - 9} \right) = x\left( {x - 9} \right)\\
\Leftrightarrow {x^3}\left( {x - 9} \right) - x\left( {x - 9} \right) = 0\\
\Leftrightarrow x\left( {x - 9} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 9 = 0\\
{x^2} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 9\\
x = 1\\
x = - 1
\end{array} \right.\\
Vay:S = \{ 0;9;1; - 1\} \\
3){x^2}\left( {x + 1} \right) + x + 1 = 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x + 1 = 0\left( {do:{x^2} + 1 > 0} \right)\\
\Leftrightarrow x = - 1\\
vay:x = - 1
\end{array}\)

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Answer 2

\(\begin{array}{l}
1)x\left( {x + 2} \right) + {x^2} = - 2{\rm{x}}\\
\Leftrightarrow x\left( {x + 2} \right) + {x^2} + 2{\rm{x}} = 0\\
\Leftrightarrow x\left( {x + 2} \right) + x\left( {x + 2} \right) = 0\\
\Leftrightarrow 2{\rm{x}}\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
vay:S = \{ 0; - 2\} \\
2){x^4} - 9{{\rm{x}}^3} = {x^2} - 9{\rm{x}}\\
\Leftrightarrow {x^3}\left( {x - 9} \right) = x\left( {x - 9} \right)\\
\Leftrightarrow {x^3}\left( {x - 9} \right) - x\left( {x - 9} \right) = 0\\
\Leftrightarrow x\left( {x - 9} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 9 = 0\\
{x^2} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 9\\
x = 1\\
x = - 1
\end{array} \right.\\
Vay:S = \{ 0;9;1; - 1\} \\
3){x^2}\left( {x + 1} \right) + x + 1 = 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x + 1 = 0\left( {do:{x^2} + 1 > 0} \right)\\
\Leftrightarrow x = - 1\\
vay:x = - 1
\end{array}\)

Answer 3

1)x(x+2)+x2=−2x⇔x(x+2)+x2+2x=0⇔x(x+2)+x(x+2)=0⇔2x(x+2)=0⇔[2x=0x+2=0⇔[x=0x=−2vay:S={0;−2}2)x4−9x3=x2−9x⇔x3(x−9)=x(x−9)⇔x3(x−9)−x(x−9)=0⇔x(x−9)(x2−1)=0⇔⎡⎢⎣x=0x−9=0x2−1=0⇔⎡⎢ ⎢ ⎢⎣x=0x=9x=1x=−1Vay:S={0;9;1;−1}3)x2(x+1)+x+1=0⇔x2(x+1)+(x+1)=0⇔(x+1)(x2+1)=0⇔x+1=0(do:x2+1>0)⇔x=−1vay:x=−1

Answer 4

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