Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a)(x - 2)(x + 5) = x(x + 4) - 38\\
\Leftrightarrow {x^2} + 5{\rm{x}} - 2{\rm{x}} - 10 = {x^2} + 4{\rm{x}} - 38\\
\Leftrightarrow 3{\rm{x}} - 10 = 4{\rm{x}} - 38\\
\Leftrightarrow 4{\rm{x}} - 3{\rm{x}} = - 10 + 38\\
\Leftrightarrow x = 28\\
Vay:x = 28\\
b){(x + 2)^2} + {(x - 1)^2} = 2(x + 3)(x + 1)\\
\Leftrightarrow {x^2} + 4{\rm{x}} + 4 + {x^2} - 2{\rm{x}} + 1 = 2\left( {{x^2} + x + 3{\rm{x}} + 3} \right)\\
\Leftrightarrow 2{{\rm{x}}^2} + 2{\rm{x}} + 5 = 2{{\rm{x}}^2} + 8{\rm{x}} + 6\\
\Leftrightarrow 2{\rm{x}} + 5 = 8{\rm{x}} + 6\\
\Leftrightarrow 8{\rm{x}} - 2{\rm{x}} = 5 - 6\\
\Leftrightarrow 6{\rm{x}} = - 1\\
\Leftrightarrow x = \frac{{ - 1}}{6}\\
Vay:S = \{ \frac{{ - 1}}{6}\} \\
c){(2x - 1)^2} + 5 = (2x - 3)(2x + 3) - x\\
\Leftrightarrow 4{{\rm{x}}^2} - 4{\rm{x}} + 1 + 5 = 4{{\rm{x}}^2} - 9 - x\\
\Leftrightarrow - 4{\rm{x}} + 6 = - x - 9\\
\Leftrightarrow 4{\rm{x}} - x = 6 + 9\\
\Leftrightarrow 3{\rm{x}} = 15\\
\Leftrightarrow x = 5\\
Vay:x = 5\\
d){(x - 1)^3} + 3{x^2} = {x^3} - 5x\\
\Leftrightarrow {x^3} - 3{{\rm{x}}^2} + 3{\rm{x}} - 1 + 3{{\rm{x}}^2} = {x^3} - 5{\rm{x}}\\
\Leftrightarrow {x^3} + 3{\rm{x}} - 1 = {x^3} - 5{\rm{x}}\\
\Leftrightarrow 3{\rm{x}} - 1 = - 5{\rm{x}}\\
\Leftrightarrow 8{\rm{x}} = 1\\
\Leftrightarrow x = \frac{1}{8}\\
Vay:x = \frac{1}{8}
\end{array}\]

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Answer 2

\[\begin{array}{l}
a)(x - 2)(x + 5) = x(x + 4) - 38\\
\Leftrightarrow {x^2} + 5{\rm{x}} - 2{\rm{x}} - 10 = {x^2} + 4{\rm{x}} - 38\\
\Leftrightarrow 3{\rm{x}} - 10 = 4{\rm{x}} - 38\\
\Leftrightarrow 4{\rm{x}} - 3{\rm{x}} = - 10 + 38\\
\Leftrightarrow x = 28\\
Vay:x = 28\\
b){(x + 2)^2} + {(x - 1)^2} = 2(x + 3)(x + 1)\\
\Leftrightarrow {x^2} + 4{\rm{x}} + 4 + {x^2} - 2{\rm{x}} + 1 = 2\left( {{x^2} + x + 3{\rm{x}} + 3} \right)\\
\Leftrightarrow 2{{\rm{x}}^2} + 2{\rm{x}} + 5 = 2{{\rm{x}}^2} + 8{\rm{x}} + 6\\
\Leftrightarrow 2{\rm{x}} + 5 = 8{\rm{x}} + 6\\
\Leftrightarrow 8{\rm{x}} - 2{\rm{x}} = 5 - 6\\
\Leftrightarrow 6{\rm{x}} = - 1\\
\Leftrightarrow x = \frac{{ - 1}}{6}\\
Vay:S = \{ \frac{{ - 1}}{6}\} \\
c){(2x - 1)^2} + 5 = (2x - 3)(2x + 3) - x\\
\Leftrightarrow 4{{\rm{x}}^2} - 4{\rm{x}} + 1 + 5 = 4{{\rm{x}}^2} - 9 - x\\
\Leftrightarrow - 4{\rm{x}} + 6 = - x - 9\\
\Leftrightarrow 4{\rm{x}} - x = 6 + 9\\
\Leftrightarrow 3{\rm{x}} = 15\\
\Leftrightarrow x = 5\\
Vay:x = 5\\
d){(x - 1)^3} + 3{x^2} = {x^3} - 5x\\
\Leftrightarrow {x^3} - 3{{\rm{x}}^2} + 3{\rm{x}} - 1 + 3{{\rm{x}}^2} = {x^3} - 5{\rm{x}}\\
\Leftrightarrow {x^3} + 3{\rm{x}} - 1 = {x^3} - 5{\rm{x}}\\
\Leftrightarrow 3{\rm{x}} - 1 = - 5{\rm{x}}\\
\Leftrightarrow 8{\rm{x}} = 1\\
\Leftrightarrow x = \frac{1}{8}\\
Vay:x = \frac{1}{8}
\end{array}\]

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