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Question

Giải phương trình

a, $\sqrt{x}+\sqrt{2-x}=2$

b, $\sqrt{1-x}+\sqrt{x}=1$
c, $\sqrt{x+5}-\sqrt{x}=1$
d, $\sqrt{x+3}+\sqrt{x+8}=5$

Lê Song Phương · Accepted Answer

The common method for the equation $\sqrt{A}+\sqrt{B}=k\left(A,B,k\ge0\right)$ (k is a constant number) usually is raise each side of the equation to the power of 2:

$\sqrt{A}+\sqrt{B}=k$ $\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2=k^2$ $\Leftrightarrow A+B+2\sqrt{AB}=k^2$$\Leftrightarrow2\sqrt{AB}=k^2-A-B$

And you raise each side of the equation to the power of 2 again: $2\sqrt{AB}=k^2-A-B\Leftrightarrow\left(2\sqrt{AB}\right)^2=\left(k^2-A-B\right)^2$ $\Leftrightarrow4AB=\left(k^2-A-B\right)^2$

And now we have eliminate all of the square roots and make it easier to solve.

But, I will give you a new method to solve this type of the equation.

a) $\sqrt{x}+\sqrt{2-x}=2$ $\left(0\le x\le2\right)$

We can easily find that $x=1$. When $x=1$, $\left\{{}\begin{matrix}\sqrt{x}=1\\sqrt{2-x}=1\end{matrix}\right.$. or $\left\{{}\begin{matrix}\sqrt{x}-1=0\\sqrt{2-x}-1=0\end{matrix}\right.$ So, we have to do something like this:

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

Notice that $\sqrt{x}+1\ne0$ and $\sqrt{2-x}+1\ne0$, we now can write the equation as below:

$\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\left(\sqrt{2-x}-1\right)\left(\sqrt{2-x}+1\right)}{\sqrt{2-x}+1}=0$

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

$\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{1-x}{\sqrt{2-x}+1}=0$

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

$\Leftrightarrow\left[{}\begin{matrix}x=1\left(take\right)\\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{2-x}+1}\left(1\right)\end{matrix}\right.$

$\left(1\right)\Leftrightarrow\sqrt{x}=\sqrt{2-x}\Leftrightarrow x=1$

Therefore, the equation a) has the root $x=1$

b)  $0\le x\le1$

Notice that $x$ can be either equal to 0 or 1

So consider $x=1$. Then, we have $\sqrt{x}=1\Leftrightarrow\sqrt{x}-1=0$ and $\sqrt{1-x}=0$. Therefore, we have to rewrite the equation like this:

$\sqrt{1-x}+\sqrt{x}=1\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\left(\sqrt{x}-1\right)=0$ $\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=0$

$\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\dfrac{x-1}{\sqrt{x}+1}=0$ $\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{1-x}}-\dfrac{1}{\sqrt{x}+1}\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}x=1\left(take\right)\\dfrac{1}{\sqrt{1-x}}=\dfrac{1}{\sqrt{x}+1}\end{matrix}\right.$or $\left[{}\begin{matrix}x=1\\sqrt{1-x}=\sqrt{x}+1\left(\cdot\right)\end{matrix}\right.$

And now, use the same method to solve $\left(\cdot\right)$

c) We have $x\ge0$

We can easily see that $x=4$, so $\sqrt{x+5}=3\Leftrightarrow\sqrt{x+5}-3=0$ and $\sqrt{x}=2\Leftrightarrow\sqrt{x}-2=0$ . Therefore, we can rewrite the equation as below:

$\sqrt{x+5}-\sqrt{x}=1\Leftrightarrow\left(\sqrt{x+5}-3\right)-\left(\sqrt{x}-2\right)=0$ $\Leftrightarrow\dfrac{\left(\sqrt{x+5}-3\right)\left(\sqrt{x+5}+3\right)}{\sqrt{x+5}+3}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=0$

$\Leftrightarrow\dfrac{\left(\sqrt{x+5}\right)^2-9}{\sqrt{x+5}+3}+\dfrac{\left(\sqrt{x}\right)^2-4}{\sqrt{x}+2}=0$

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

$\Leftrightarrow\left(x-4\right)\left(\dfrac{1}{\sqrt{x+5}+3}+\dfrac{1}{\sqrt{x}+2}\right)=0$

$\Leftrightarrow...$

Notice that $\dfrac{1}{\sqrt{x+5}+3}+\dfrac{1}{\sqrt{x}+2}$ can't be equal to 0. So this equation only have the root $x=4$

d) Similar to the equations above.

Câu trả lời: The common method for the equation $\sqrt{A}+\sqrt{B}=k\left(A,B,k\ge0\right)$ (k is a constant number) usually is raise each side of the equation to the power of 2: