Toán Lớp 8: giúp mình giải bài này với 2) (x-3) (x+3) -(x-2)^2 =1 3) (x-1) (x^2 +x+1) – x(x^2+2)-4=0 4) x(x-5) (x+5) – (x+2 )(x^2-2x+4 )=3
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Toán Lớp 8: giúp mình giải bài này với 2) (x-3) (x+3) -(x-2)^2 =1 3) (x-1) (x^2 +x+1) – x(x^2+2)-4=0 4) x(x-5) (x+5) – (x+2 )(x^2-2x+4 )=3
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2)\left( {x – 3} \right)\left( {x + 3} \right) – {\left( {x – 2} \right)^2} = 1\\
\Leftrightarrow {x^2} – 9 – \left( {{x^2} – 4x + 4} \right) = 1\\
\Leftrightarrow {x^2} – 9 – {x^2} + 4x – 4 = 1\\
\Leftrightarrow 4x = 14\\
\Leftrightarrow x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
3)\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) – x\left( {{x^2} + 2} \right) – 4 = 0\\
\Leftrightarrow {x^3} – 1 – {x^3} – 2x – 4 = 0\\
\Leftrightarrow 2x = – 5\\
\Leftrightarrow x = – \dfrac{{52}}{{}}\\
Vậy\,x = – \dfrac{5}{2}\\
4)x\left( {x – 5} \right)\left( {x + 5} \right) – \left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) = 3\\
\Leftrightarrow x\left( {{x^2} – 25} \right) – \left( {{x^3} + {2^3}} \right) = 3\\
\Leftrightarrow {x^3} – 25x – {x^3} – 8 = 3\\
\Leftrightarrow 25x = – 11\\
\Leftrightarrow x = – \dfrac{{11}}{{25}}\\
Vậy\,x = – \dfrac{{11}}{{25}}
\end{array}$