Đặt \(t=\sqrt{3x^2+5x+1}\left(t\ge0\right)\)

pt đã cho trở thành: \(\sqrt{t^2+7}-t=1\Leftrightarrow\sqrt{t^2+7}=t+1\)

- bình phương 2 vế, giải ra t, trả lại nghiệm x, tìm x

\(\sqrt {3{x^2} + 5x + 8} - \sqrt {3{x^2} + 5x + 1} = 1\\ \text{Điều kiện}: \forall x \in \mathbb{R}\\ \text{Đặt}:\sqrt {3{x^2} + 5x + 8} =a; \sqrt {3{x^2} + 5x + 1} = b\\ \Rightarrow \left\{ \begin{array}{l} a - b = 1\\ {a^2} - {b^2} = 7 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = b + 1\\ {\left( {b + 1} \right)^2} - {b^2} = 7 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = b + 1\\ 2b = 6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 4\\ b = 3 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \sqrt {3{x^2} + 5x + 8} = 4\\ \sqrt {3{x^2} + 5x + 1} = 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3{x^2} + 5x + 8 = 16\\ 3{x^2} + 5x + 1 = 9 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3{x^2} + 5x - 8 = 0\\ 3{x^2} + 5x - 8 = 0 \end{array} \right.\\ \Leftrightarrow \left( {x - 1} \right)\left( {3x + 8} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - \dfrac{8}{3} \end{array} \right. \)

\(\text{Cách khác:}\\ \text{Đặt}:t = \sqrt {3{x^2} + 5x + 8} \left( {t \ge 0} \right) \text{thì} t^2=3x^2+5x+8\\ PT:t-\sqrt{t^2-7}=1 \Leftrightarrow \sqrt{t^2-7} = t- 1 \)

\( \Leftrightarrow \left\{ \begin{array}{l} t - 1 \ge 0\\ {t^2} - 7 = {\left( {t - 1} \right)^2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t \ge 1\\ 2t = 8 \end{array} \right. \Leftrightarrow t = 4\)

\(\text{Do đó:}\) \(\sqrt{3x^2+5x+8}=4\Leftrightarrow3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\Leftrightarrow\)\(x=1 \text{hoặc} x =- \dfrac{8}{3}\)

a) Đặt \(a=\sqrt[3]{1+\sqrt{x}};b=\sqrt[3]{1-\sqrt{x}}\)

\(\Rightarrow a^3+b^3=2\) kết hợp với đề bài

\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a^2-ab+b^2\right)=2\\a+b=2\end{matrix}\right.\)

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b và c tương tự

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)

Câu a bạn coi lại đề

b. ĐKXĐ: \(x\ge0;x\ne1\)

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

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

\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)

\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )

\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)

\(\Leftrightarrow20x^2+16x-1=0\)

\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)

Bạn xem lại đề câu a.

a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)

đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)

\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)

\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)

pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)

\(\Leftrightarrow t^2-2t-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)

suy ra tìm đc x

câu b đặt t =\(3x^2+5x+8\)

ta có pt \(\Leftrightarrow\sqrt{t}-\sqrt{t-7}=1\)

\(\Rightarrow t=16\)

\(\Leftrightarrow3x^2+5x+8=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{3}\end{matrix}\right.\)