1) (x-1)³+3(x-3)²-(x+2)(x²-2x+4)=(x+2)³-(x-3)(x²+9)-6x²+5
2) (x-5)(x+5)-(x-2)³-7x²+(x+1)(x²-x+1)=(x+3)³-(x³+9x²)
3) (x-4)³-(x+5)(x²+5x+25)=(x+2)(x-2x+4)-(x+4)³-(x-7)
4) (x+1)³-(x+3)(x²-3x+9)=(x-3)³+3(2x+1)²-(x³-5x+1)
1) (x-1)³+3(x-3)²-(x+2)(x²-2x+4)=(x+2)³-(x-3)(x²+9)-6x²+5 2) (x-5)(x+5)-(x-2)³-7x²+(x+1)(x²-x+1)=(x+3)³-(x³+9x²) 3) (x-4)³-(x+5)(x²+5x+25)=(x+2)(x-2x+
By Cora
Đáp án:
Giải thích các bước giải:
Đáp án: 1)$S=\{\dfrac{{ – 9 + \sqrt {15} }}{3};\dfrac{{ – 9 – \sqrt {15} }}{3}\}$
2)$S=\{\dfrac{{ – 43}}{{39}}\}$
4)$S=\{\dfrac{1}{{41}}\}$
Giải thích các bước giải:
$\begin{array}{l}
1){\left( {x – 1} \right)^3} + 3{\left( {x – 3} \right)^2} – \left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) = {\left( {x + 2} \right)^3} – \left( {x – 3} \right)\left( {{x^2} + 9} \right) – 6{x^2} + 5\\
\Leftrightarrow {\left( {x – 1} \right)^3} + 3{\left( {x – 3} \right)^2} – \left( {{x^3} + 8} \right) – {\left( {x + 2} \right)^3} + \left( {x – 3} \right)\left( {{x^2} + 9} \right) + 6{x^2} – 5\\
\Leftrightarrow \left( { – 3} \right)\left[ {{{\left( {x – 1} \right)}^2} + \left( {x – 1} \right)\left( {x + 2} \right) + {{\left( {x + 2} \right)}^2}} \right] + \left( {x – 3} \right)\left[ {3\left( {x – 3} \right) + {x^2} + 9} \right] – {x^3} + 6{x^2} – 13 = 0\\
\Leftrightarrow \left( { – 3} \right)\left( {3{x^2} + 3x + 3} \right) + \left( {x – 3} \right)\left( {{x^2} + 3x} \right) – {x^3} + 6{x^2} – 13 = 0\\
\Leftrightarrow 3{x^2} + 18x + 22 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ – 9 + \sqrt {15} }}{3}\\
x = \dfrac{{ – 9 – \sqrt {15} }}{3}
\end{array} \right.
\end{array}$
Vậy phương trình có tập nghiệm là: $S=\{\dfrac{{ – 9 + \sqrt {15} }}{3};\dfrac{{ – 9 – \sqrt {15} }}{3}\}$
$\begin{array}{l}
2)\left( {x – 5} \right)\left( {x + 5} \right) – {\left( {x – 2} \right)^3} – 7{x^2} + \left( {x + 1} \right)\left( {{x^2} – x + 1} \right) = {\left( {x + 3} \right)^3} – \left( {{x^3} + 9{x^2}} \right)\\
\Leftrightarrow {x^2} – 25 – {\left( {x – 2} \right)^3} – {\left( {x + 3} \right)^3} – 7{x^2} + {x^3} + 1 + {x^3} + 9{x^2} = 0\\
\Leftrightarrow 2{x^3} + 3{x^2} – 24 – \left( {2x + 1} \right)\left[ {{{\left( {x – 2} \right)}^2} – \left( {x – 2} \right)\left( {x + 3} \right) + {{\left( {x + 3} \right)}^2}} \right] = 0\\
\Leftrightarrow 2{x^3} + 3{x^2} – 24 – \left( {2x + 1} \right)\left( {{x^2} + x + 19} \right) = 0\\
\Leftrightarrow – 39x + 43 \Leftrightarrow x = \dfrac{{ – 43}}{{39}}
\end{array}$
Vậy phương trình có tập nghiệm là: $S=\{\dfrac{{ – 43}}{{39}}\}$
3) Bạn kiểm tra lại đề bài.
$\begin{array}{l}
4){\left( {x + 1} \right)^3} – \left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) = {\left( {x – 3} \right)^3} + 3{\left( {2x + 1} \right)^2} – \left( {{x^3} – 5x + 1} \right)\\
\Leftrightarrow {\left( {x + 1} \right)^3} – {\left( {x – 3} \right)^3} – {x^3} – 27 – 3\left( {4{x^2} + 4x + 1} \right) + \left( {{x^3} – 5x + 1} \right) = 0\\
\Leftrightarrow 4\left[ {{{\left( {x + 1} \right)}^2} + \left( {x + 1} \right)\left( {x – 3} \right) + {{\left( {x – 3} \right)}^2}} \right] – 12{x^2} – 17x – 29 = 0\\
\Leftrightarrow 4\left[ {3{x^2} – 6x + 7} \right] – 12{x^2} – 17x – 29 = 0\\
\Leftrightarrow – 41x – 1 = 0 \Leftrightarrow x = \dfrac{1}{{41}}
\end{array}$
Vậy phương trình có tập nghiệm là: $S=\{\dfrac{1}{{41}}\}$