Cho A=3(2x-30(3x+2)-4(x+4)(4x-3)+9x(4-x). Để A có giá trị bằng 0 thì X??? 17/08/2021 Bởi Alaia Cho A=3(2x-30(3x+2)-4(x+4)(4x-3)+9x(4-x). Để A có giá trị bằng 0 thì X???
Đáp án: \(\left[ \begin{array}{l}x = \left( {\sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\x = \left( { – \sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}A = 3\left( {2x – 3} \right)\left( {3x + 2} \right) – 4\left( {x + 4} \right)\left( {4x – 3} \right) + 9x\left( {4 – x} \right)\\ = \left( {6x – 9} \right)\left( {3x + 2} \right) – \left( {4x + 16} \right)\left( {4x – 3} \right) + 26x – 9{x^2}\\ = 18{x^2} + 12x – 27x – 18 – 16{x^2} + 12x – 64x + 48 + 26x – 9{x^2}\\ = – 7{x^2} – 41x + 30\\A = 0\\ \to – 7{x^2} – 41x + 30 = 0\\ \to – \left( {7{x^2} + 41x – 30} \right) = 0\\ \to – \left[ {{{\left( {x\sqrt 7 } \right)}^2} + 2.x\sqrt 7 .\dfrac{{41\sqrt 7 }}{{14}} + {{\left( {\dfrac{{41\sqrt 7 }}{{14}}} \right)}^2} – \dfrac{{2521}}{{28}}} \right] = 0\\ \to – {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} + \dfrac{{2521}}{{28}} = 0\\ \to {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} = \dfrac{{2521}}{{28}}\\ \to \left[ \begin{array}{l}x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = \sqrt {\dfrac{{2521}}{{28}}} \\x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = – \sqrt {\dfrac{{2521}}{{28}}} \end{array} \right.\\ \to \left[ \begin{array}{l}x = \left( {\sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\x = \left( { – \sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = \left( {\sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\
x = \left( { – \sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 3\left( {2x – 3} \right)\left( {3x + 2} \right) – 4\left( {x + 4} \right)\left( {4x – 3} \right) + 9x\left( {4 – x} \right)\\
= \left( {6x – 9} \right)\left( {3x + 2} \right) – \left( {4x + 16} \right)\left( {4x – 3} \right) + 26x – 9{x^2}\\
= 18{x^2} + 12x – 27x – 18 – 16{x^2} + 12x – 64x + 48 + 26x – 9{x^2}\\
= – 7{x^2} – 41x + 30\\
A = 0\\
\to – 7{x^2} – 41x + 30 = 0\\
\to – \left( {7{x^2} + 41x – 30} \right) = 0\\
\to – \left[ {{{\left( {x\sqrt 7 } \right)}^2} + 2.x\sqrt 7 .\dfrac{{41\sqrt 7 }}{{14}} + {{\left( {\dfrac{{41\sqrt 7 }}{{14}}} \right)}^2} – \dfrac{{2521}}{{28}}} \right] = 0\\
\to – {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} + \dfrac{{2521}}{{28}} = 0\\
\to {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} = \dfrac{{2521}}{{28}}\\
\to \left[ \begin{array}{l}
x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = \sqrt {\dfrac{{2521}}{{28}}} \\
x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = – \sqrt {\dfrac{{2521}}{{28}}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \left( {\sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\
x = \left( { – \sqrt {\dfrac{{2521}}{{28}}} – \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7
\end{array} \right.
\end{array}\)