Câu hỏi của Ngọc Băng

Question

giải pt

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

2.$\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}$

3.$\sqrt{14-x}-\sqrt{x-4}\sqrt{x-1}$

4. \(3+\sqrt{x+2\sqrt{...

qwerty · Accepted Answer

2. $\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}$ (2)

$\Leftrightarrow\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}-\dfrac{7}{\sqrt{x-3}}=0$

$\Leftrightarrow\dfrac{\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7}{\sqrt{x-3}}=0$

$\Leftrightarrow\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7=0$

$\Leftrightarrow\left|x\right|-16+\sqrt{x^2-9}-7=0$

$\Leftrightarrow\left|x\right|-23+\sqrt{x^2-9}=0$

$\Leftrightarrow\sqrt{x^2-9}=-\left|x\right|+23$

$\Leftrightarrow x^2-9=-\left(-\left|x\right|+23\right)^2$

$\Leftrightarrow x^2-9=-\left(-\left|x\right|\right)^2-46\cdot\left|x\right|+529$

$\Leftrightarrow x^2-9=\left|x\right|^2-46+\left|x\right|+529$

$\Leftrightarrow x^2-9=x^2-46\cdot\left|x\right|+529$

$\Leftrightarrow-9=-46\cdot\left|x\right|+529$

$\Leftrightarrow46\cdot\left|x\right|=529+9$

$\Leftrightarrow49\cdot\left|x\right|=538$

$\Leftrightarrow\left|x\right|=\dfrac{269}{23}$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{269}{23}\x=-\dfrac{269}{23}\end{matrix}\right.$

Sau khi dùng phép thử ta nhận thấy $x\ne-\dfrac{269}{23}$

Vậy tập nghiệm phương trình (1) là $S=\left\{\dfrac{269}{23}\right\}$

3. sửa đề: $\sqrt{14-x}=\sqrt{x-4}\sqrt{x-1}$ (3)

$\Leftrightarrow\sqrt{14-x}=\sqrt{\left(x-4\right)\left(x-1\right)}$

$\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-x-4x+4}$

$\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-5x+4}$

$\Leftrightarrow14-x=x^2-5x+4$

$\Leftrightarrow14-x-x^2+5x-4=0$

$\Leftrightarrow10+4x-x^2=0$

$\Leftrightarrow-x^2+4x+10=0$

$\Leftrightarrow x^2-4x-10=0$

$\Leftrightarrow x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot\left(-10\right)}}{2\cdot1}$

$\Leftrightarrow x=\dfrac{4\pm\sqrt{16+40}}{2}$

$\Leftrightarrow x=\dfrac{4\pm\sqrt{56}}{2}$

$\Leftrightarrow x=\dfrac{4\pm2\sqrt{14}}{2}$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{14}}{2}\x=\dfrac{4+2\sqrt{14}}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{14}\x=2-\sqrt{14}\end{matrix}\right.$

sau khi dùng phép thử ta nhận thấy $x\ne2-\sqrt{14}$

Vậy tập nghiệm phương trình (3) là $S=\left\{2+\sqrt{14}\right\}$Câu trả lời: 2. $\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}$ (2)