ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\sqrt{x+3}-\sqrt{2x-1}=\sqrt{3x-2}\)

\(\Leftrightarrow\sqrt{x+3}=\sqrt{2x-1}+\sqrt{3x-2}\)

\(\Leftrightarrow x+3=2x-1+3x-2+2\sqrt{\left(2x-1\right)\left(3x-2\right)}\)

\(\Leftrightarrow3-2x=\sqrt{\left(2x-1\right)\left(3x-2\right)}\) (\(x\le\frac{3}{2}\))

\(\Leftrightarrow\left(3-2x\right)^2=\left(2x-1\right)\left(3x-2\right)\)

\(\Leftrightarrow4x^2-12x+9=6x^2-7x+2\)

\(\Leftrightarrow2x^2+5x-7=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{7}{2}< \frac{2}{3}\left(l\right)\end{matrix}\right.\)

ĐKXĐ: x>=-1/2

\(\sqrt[3]{x-3}+3\sqrt{2x+1}=10\)

=>\(\sqrt[3]{x-3}-1+3\sqrt{2x+1}-9=0\)

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

=>\(\dfrac{x-4}{\sqrt[3]{\left(x-3\right)^2}+\sqrt[3]{x-3}+1}+3\cdot\dfrac{2x+1-9}{\sqrt{2x+1}+3}=0\)

=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt[3]{\left(x-3\right)^2}+\sqrt[3]{x-3}+1}+\dfrac{6}{\sqrt{2x+1}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

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

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

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

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

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

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

\(\Rightarrow x-1=0\Rightarrow x=1\)

cảm ơn bạn ace legona nhá

Điều kiện $x\geq 1$.

• Nếu x>2 thì VT>6>VP
• Nếu x<2 thì VT<6<VP

Vậy phương trình có nghiệm duy nhất x=2

4885

ở dưới ghi - ở trên ghi + rốt cuộc đề nào đúng

Dạ 2 đề là 1 ạ tại em muốn ghi lại cho mọi người hiểu ạ