giải PT : $\sqrt[3]{2x+3}$ – $\sqrt[3]{x-1}$ +x+4=0 02/07/2021 Bởi Faith giải PT : $\sqrt[3]{2x+3}$ – $\sqrt[3]{x-1}$ +x+4=0
Đáp án: đặt 2x+3=a x-1=b => $\sqrt[3]{a}$ -$\sqrt[3]{b}$ +a-b=0h các bước giải: <=>$\sqrt[3]{a}$ -$\sqrt[3]{b}$ +( $\sqrt[3]{a}$ -$\sqrt[3]{b})(a+$\sqrt[]{a}$ $\sqrt[]{b}$ +b) <=> ($\sqrt[3]{a}$ -$\sqrt[3]{b}$)(a+$\sqrt[]{a}$ *$\sqrt[]{b}$ +b+1)=0 <=> $\sqrt[3]{a}$ -$\sqrt[3]{b}$=0 <=> $\sqrt[3]{a}$ =$\sqrt[3]{b}$ <=> a=b <=> 2x+3=x-1 <=> x=-4 vậy x = -4 Bình luận
$\begin{array}{l} \sqrt[3]{{2x + 3}} – \sqrt[3]{{x – 1}} + x + 4 = 0\\ \Leftrightarrow \dfrac{{2x + 3 – x + 1}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + x + 4 = 0\\ \Leftrightarrow \dfrac{{x + 4}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + x + 4 = 0\\ \Leftrightarrow \left( {x + 4} \right)\left[ {\dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 4 = 0 \Leftrightarrow x = – 4\\ \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1 = 0\left( * \right) \end{array} \right.\\ \left( * \right) \Rightarrow \sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}} = {a^2} + ab + {b^2}\left( {a = \sqrt[3]{{2x + 3}},b = \sqrt[3]{{x – 1}}} \right)\\ \Rightarrow {a^2} + ab + {b^2} > 0\forall a,b \in R\\ \Rightarrow \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1 > 1\\ \Rightarrow \left( * \right)VN\\ \Rightarrow S = \left\{ { – 4} \right\} \end{array}$ Bình luận
Đáp án:
đặt 2x+3=a
x-1=b
=> $\sqrt[3]{a}$ -$\sqrt[3]{b}$ +a-b=0h các bước giải:
<=>$\sqrt[3]{a}$ -$\sqrt[3]{b}$ +( $\sqrt[3]{a}$ -$\sqrt[3]{b})(a+$\sqrt[]{a}$ $\sqrt[]{b}$ +b)
<=> ($\sqrt[3]{a}$ -$\sqrt[3]{b}$)(a+$\sqrt[]{a}$ *$\sqrt[]{b}$ +b+1)=0
<=> $\sqrt[3]{a}$ -$\sqrt[3]{b}$=0
<=> $\sqrt[3]{a}$ =$\sqrt[3]{b}$
<=> a=b
<=> 2x+3=x-1
<=> x=-4
vậy x = -4
$\begin{array}{l} \sqrt[3]{{2x + 3}} – \sqrt[3]{{x – 1}} + x + 4 = 0\\ \Leftrightarrow \dfrac{{2x + 3 – x + 1}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + x + 4 = 0\\ \Leftrightarrow \dfrac{{x + 4}}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + x + 4 = 0\\ \Leftrightarrow \left( {x + 4} \right)\left[ {\dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 4 = 0 \Leftrightarrow x = – 4\\ \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1 = 0\left( * \right) \end{array} \right.\\ \left( * \right) \Rightarrow \sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}} = {a^2} + ab + {b^2}\left( {a = \sqrt[3]{{2x + 3}},b = \sqrt[3]{{x – 1}}} \right)\\ \Rightarrow {a^2} + ab + {b^2} > 0\forall a,b \in R\\ \Rightarrow \dfrac{1}{{\sqrt[3]{{{{\left( {2x + 3} \right)}^2}}} + \sqrt[3]{{\left( {2x + 3} \right)\left( {x – 1} \right)}} + \sqrt[3]{{{{\left( {x – 1} \right)}^2}}}}} + 1 > 1\\ \Rightarrow \left( * \right)VN\\ \Rightarrow S = \left\{ { – 4} \right\} \end{array}$