Question

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Phương pháp 2. Biến đổi về phương trình tích

a \(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

b \(2\sqrt[3]{\left(x+3\right)^2}-\sqrt[3]{\left(x-3\right)^2}=\sqrt[3]{x^2-9}\)

c \(\sqrt{2x+1}+3\sqrt{4x^2-2x+1}=3+\sqrt{8x^3+1}\)

d \(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)

Answer 1

[𝓓𝓾𝔂 𝓐𝓷𝓱]

a) ĐK: \(x\ge3\)

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

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

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

     \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)

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Answer 2

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