Question

[SA Na]

giải giúp mình các pt sau đây nha
1. \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)

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

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

4. \(4x^2-x+4=3x\sqrt{x+\dfrac{1}{x}}\)

5. \(\sqrt[4]{x^2+x+1}+\sqrt[4]{x^2-x+1}=2\sqrt[4]{x}\)

6. \(4x^2-3x-4=\sqrt[3]{x^4-x^2}\)

giải nhanh giúp mình nha
thanks trước

Answer 1

[Tuyển Trần Thị]

a,dk x>0

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

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

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

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

kh vs dé bài ta có hệ \(\left\{{}\begin{matrix}\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\\\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\end{matrix}\right.\)

cộng vs nhau ta có

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

\(\Leftrightarrow3\sqrt{2x^2+x+1}=5x+1\)

giải ra ta có x=1(tm) x=-8/7 (l)

Answer 2

[Tuyển Trần Thị]

b, dk tu xd nhé

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

\(\Leftrightarrow2x\left(\dfrac{1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1\left(l\right)\end{matrix}\right.\)

ns \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}>1\)

\(\Rightarrow x=0\left(tm\right)\)