Đáp án:
\(\begin{array}{l}
a,\\
x = \dfrac{{15}}{8}\\
b,\\
x = \dfrac{3}{4}\\
c,\\
x = 1\\
d,\\
x = \dfrac{{34}}{{15}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x - 5} \right)\left( {x - 4} \right) - \left( {x + 1} \right)\left( {x - 2} \right) = 7\\
\Leftrightarrow \left( {{x^2} - 4x - 5x + 20} \right) - \left( {{x^2} - 2x + x - 2} \right) = 7\\
\Leftrightarrow \left( {{x^2} - 9x + 20} \right) - \left( {{x^2} - x - 2} \right) = 7\\
\Leftrightarrow {x^2} - 9x + 20 - {x^2} + x + 2 - 7 = 0\\
\Leftrightarrow - 8x + 15 = 0\\
\Leftrightarrow 8x = 15\\
\Leftrightarrow x = \dfrac{{15}}{8}\\
b,\\
5x\left( {x - 3} \right) = \left( {x - 2} \right)\left( {5x - 1} \right) - 5\\
\Leftrightarrow 5{x^2} - 15x = \left( {5{x^2} - x - 10x + 2} \right) - 5\\
\Leftrightarrow 5{x^2} - 15x = 5{x^2} - 11x - 3\\
\Leftrightarrow 5{x^2} - 15x - 5{x^2} + 11x + 3 = 0\\
\Leftrightarrow - 4x + 3 = 0\\
\Leftrightarrow 4x = 3\\
\Leftrightarrow x = \dfrac{3}{4}\\
c,\\
\left( {x - 5} \right)\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 2} \right)\\
\Leftrightarrow \left( {x - 5} \right)\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right).\left[ {\left( {x - 5} \right) - \left( {x - 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right).\left( {x - 5 - x + 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right).\left( { - 3} \right) = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
d,\\
6\left( {x - 3} \right)\left( {x - 4} \right) - 6x\left( {x - 2} \right) = 4\\
\Leftrightarrow 6.\left( {{x^2} - 4x - 3x + 12} \right) - \left( {6{x^2} - 12x} \right) = 4\\
\Leftrightarrow 6\left( {{x^2} - 7x + 12} \right) - \left( {6{x^2} - 12x} \right) = 4\\
\Leftrightarrow 6{x^2} - 42x + 72 - 6{x^2} + 12x - 4 = 0\\
\Leftrightarrow - 30x + 68 = 0\\
\Leftrightarrow 30x - 68 = 0\\
\Leftrightarrow 30x = 68\\
\Leftrightarrow x = \dfrac{{34}}{{15}}
\end{array}\)
