Question

tìm x

1. 4x-\(\sqrt{9x^2+6x+1}\)=0

2. \(\sqrt{4x}+20-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=0\)

3. \(\sqrt{x^2}+x+1=x+2\)

4. \(\sqrt{6x^2}+2-\sqrt{3x^2}=\sqrt{26}-\sqrt{13}\)

5. \(\dfrac{\sqrt{x}-2}{\sqrt{x}-4}=\dfrac{\sqrt{x}-6}{\sqrt{x}-7}\)

6. \(\sqrt{4x^2-4x+1}=3\)

giải từng bước ra cho mik nha,thank mn

Answer 1

1. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow 4x=\sqrt{(3x+1)^2}$

\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (4x)^2=(3x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (4x-3x-1)(4x+3x+1)=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (x-1)(7x+1)=0\end{matrix}\right.\Leftrightarrow x=1\)

Vậy $x=1$ là nghiệm của pt.

 

Answer 2

2. ĐKXĐ: $x\geq -5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x+5}-3\sqrt{5+x}+\frac{4}{3}.\sqrt{9}.\sqrt{x+5}=0$

$\Leftrightarrow 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=0$

$\Leftrightarrow 3\sqrt{x+5}=0$

$\Leftrightarrow \sqrt{x+5}=0$

$\Leftrightarrow x=-5$

 

Answer 3

3. 

$\sqrt{x^2+x+1}=x+2$

\(\Leftrightarrow \left\{\begin{matrix} x+2\geq 0\\ x^2+x+1=(x+2)^2=x^2+4x+4\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 3x=-3\end{matrix}\right.\Leftrightarrow x=-1\)

Vậy $x=-1$ là nghiệm của pt.

 

Answer 4

4.

$\sqrt{6x^2}+2-\sqrt{3x^2}=\sqrt{26}-\sqrt{13}$

$\Leftrihgtarrow \sqrt{6}|x|-\sqrt{3}|x|=\sqrt{26}-\sqrt{13}-2$

$\Leftrightarrow |x|(\sqrt{6}-\sqrt{3})=\sqrt{26}-\sqrt{13}-2< 0$

$\Rightarrow |x|< 0$ (vô lý)

Vậy pt vô nghiệm.

 

 

Answer 5

5. ĐKXĐ: $x\geq 0; x\neq 16; x\neq 49$

PT \(\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-7)=(\sqrt{x}-4)(\sqrt{x}-6)\)

\(\Leftrightarrow x-9\sqrt{x}+14=x-10\sqrt{x}+24\)

\(\Leftrightarrow \sqrt{x}=10\Leftrightarrow x=100\) (thỏa mãn)

 

Answer 6

6.

$\sqrt{4x^2-4x+1}=3$

$\Leftrightarrow \sqrt{(2x-1)^2}=3$
$\Leftrightarrow |2x-1|=3$

$\Leftrightarrow 2x-1=3$ hoặc $2x-1=-3$

$\Leftrightarrow x=2$ hoặc $x=-1$ (đều thỏa mãn)