Câu hỏi của Chiều Xuân

Question

giúp mình với

căn bậc 3 (2x-1) + căn bậc 3 (x -1 ) =1

Lightning Farron · Accepted Answer

$\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1$

$\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0$

$\Leftrightarrow\dfrac{2x-1-1}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\sqrt[3]{\left(x-1\right)}^2}=0$

$\Leftrightarrow\dfrac{2\left(x-1\right)}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\sqrt[3]{\left(x-1\right)}^2}=0$

$\Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{1}{\sqrt[3]{\left(x-1\right)}^2}\right)=0$

Dễ thấy: $\dfrac{2}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{1}{\sqrt[3]{\left(x-1\right)}^2}>0$

$\Rightarrow x-1=0\Rightarrow x=1$Câu trả lời: $\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1$

$\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0$

$\Leftrightarrow\dfrac{2x-1-1}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\sqrt[3]{\left(x-1\right)}^2}=0$

$\Leftrightarrow\dfrac{2\left(x-1\right)}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\sqrt[3]{\left(x-1\right)}^2}=0$

$\Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{1}{\sqrt[3]{\left(x-1\right)}^2}\right)=0$

Dễ thấy: $\dfrac{2}{\sqrt[3]{\left(2x-1\right)}^2+2\sqrt[3]{2x-1}+1}+\dfrac{1}{\sqrt[3]{\left(x-1\right)}^2}>0$

$\Rightarrow x-1=0\Rightarrow x=1$

Chiều Xuân · Answer

cảm ơn bạn ace legona nháCâu trả lời: cảm ơn bạn ace legona nhá