Question

Giúp em bài 3

Answer 1

Bài 3:

c. \(\sqrt{x^2+9}+3=2x\)

<=> \(\sqrt{x^2+9}=2x-3\)

<=> x² + 9 = (2x - 3)²

<=> x² + 9 = 4x² - 12x + 9

<=> 9 - 9 = 4x² - x² - 12x

<=> 3x² - 12x = 0

<=> 3x(x - 4) = 0

<=> \(\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Answer 2

Bài 3:

a. ĐKXĐ: $x\geq \frac{-3}{5}$

PT $\Leftrightarrow 3\sqrt{4(5x+3)}-\frac{2}{3}\sqrt{9(5x+3)}=2$

$\Leftrightarrow 3.\sqrt{4}.\sqrt{5x+3}-\frac{2}{3}.\sqrt{9}.\sqrt{5x+3}=2$

$\Leftrightarrow 6\sqrt{5x+3}-2\sqrt{5x+3}=2$

$\Leftrightarrow 4\sqrt{5x+3}=2$

$\Leftrightarrow \sqrt{5x+3}=\frac{1}{2}$

$\Leftrightarrow 5x+3=\frac{1}{4}$

$\Leftrightarrow x=\frac{-11}{20}$ (tm)

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{4(x-3)}-6\sqrt{\frac{1}{4}(x-3)}=10-\sqrt{9(x-3)}$

$\Leftrightarrow 2\sqrt{x-3}-3\sqrt{x-3}=10-3\sqrt{x-3}$

$\Leftrightarrow 2\sqrt{x-3}=10$

$\Leftrightarrow \sqrt{x-3}=5$

$\Leftrightarrow x-3=25$

$\Leftrightarrow x=28$ (tm)

c.

PT $\Leftrightarrow \sqrt{x^2+9}=2x-3$
\(\Rightarrow \left\{\begin{matrix} 2x-3\geq 0\\ x^2+9=(2x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{3}{2}\\ 3x^2-12x=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{3}{2}\\ 3x(x-4)=0\end{matrix}\right.\)

$\Leftrightarrow x=4$ 

d. PT \(\Rightarrow \left\{\begin{matrix} 1-5x\geq 0\\ 9x^2-6x+1=(1-5x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{5}\\ 16x^2-4x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{5}\\ 4x(4x-1)=0\end{matrix}\right.\Leftrightarrow x=0\)