1/X^3 -(2+3x)(3x-2)= (4-2x +x^2)(x+2)-(3x+5) Tìm x 2/ chứng minh biểu thức 2+4x^2 – 3x>0 Giúp mình với ạ đáng gấp ạ 17/07/2021 Bởi Caroline 1/X^3 -(2+3x)(3x-2)= (4-2x +x^2)(x+2)-(3x+5) Tìm x 2/ chứng minh biểu thức 2+4x^2 – 3x>0 Giúp mình với ạ đáng gấp ạ
Đáp án: $\begin{array}{l}1)\\{x^3} – \left( {2 + 3x} \right)\left( {3x – 2} \right) = \left( {4 – 2x + {x^2}} \right)\left( {x + 2} \right)\\ – \left( {3x + 5} \right)\\ \Rightarrow {x^3} – \left( {9{x^2} – 4} \right) = {x^3} + 8 – 3x – 5\\ \Rightarrow {x^3} – 9{x^2} + 4 = {x^3} – 3x + 3\\ \Rightarrow 9{x^2} – 3x – 1 = 0\\ \Rightarrow {\left( {3x} \right)^2} – 2.3x.\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{5}{4}\\ \Rightarrow {\left( {3x – \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\\ \Rightarrow \left[ \begin{array}{l}3x – \dfrac{1}{2} = \dfrac{{\sqrt 5 }}{2}\\3x – \dfrac{1}{2} = – \dfrac{{\sqrt 5 }}{2}\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{{3\sqrt 5 + 3}}{2}\\x = \dfrac{{ – 3\sqrt 5 + 3}}{2}\end{array} \right.\\2)\\2 + 4{x^2} – 3x\\ = 4{x^2} – 3x + 2\\ = {\left( {2x} \right)^2} – 2.2x.\dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{23}}{{16}}\\ = {\left( {2x – \dfrac{3}{4}} \right)^2} + \dfrac{{23}}{{16}} \ge \dfrac{{23}}{{16}} > 0\\Vay\,2 + 4{x^2} – 3x > 0\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
1)\\
{x^3} – \left( {2 + 3x} \right)\left( {3x – 2} \right) = \left( {4 – 2x + {x^2}} \right)\left( {x + 2} \right)\\
– \left( {3x + 5} \right)\\
\Rightarrow {x^3} – \left( {9{x^2} – 4} \right) = {x^3} + 8 – 3x – 5\\
\Rightarrow {x^3} – 9{x^2} + 4 = {x^3} – 3x + 3\\
\Rightarrow 9{x^2} – 3x – 1 = 0\\
\Rightarrow {\left( {3x} \right)^2} – 2.3x.\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{5}{4}\\
\Rightarrow {\left( {3x – \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\\
\Rightarrow \left[ \begin{array}{l}
3x – \dfrac{1}{2} = \dfrac{{\sqrt 5 }}{2}\\
3x – \dfrac{1}{2} = – \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{3\sqrt 5 + 3}}{2}\\
x = \dfrac{{ – 3\sqrt 5 + 3}}{2}
\end{array} \right.\\
2)\\
2 + 4{x^2} – 3x\\
= 4{x^2} – 3x + 2\\
= {\left( {2x} \right)^2} – 2.2x.\dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{23}}{{16}}\\
= {\left( {2x – \dfrac{3}{4}} \right)^2} + \dfrac{{23}}{{16}} \ge \dfrac{{23}}{{16}} > 0\\
Vay\,2 + 4{x^2} – 3x > 0
\end{array}$