Lưu ý là \(0\le n\le360\) nhé.

C1: Giải bằng cách tính delta

C2: PT có dạng a-b+c=0

C3: Nhẩm nghiệm có 1 nghiệm là -1 dùng phép chia đ thức

\(PT\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)

C1: 2x²-4x-6=0
⇔2(x²-2x-3)=0
⇔x²-2x-3=0
⇔x²+x-3x-3=0
⇔x(x+1) - 3(x+1)=0
⇔(x+1)(x-3)=0
⇔ \(_{\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\)

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

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.

Ý a, chỗ ( x-1 ) \(\dfrac{1}{\sqrt{x}+1}\) - \(\dfrac{1}{\sqrt{2-x}+1}\) = 0 tại sao lại làm mất được (x-1) vậy ạ ? 

Lời giải:

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

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

$(x+1)^2=4-2\sqrt{3}=(\sqrt{3}-1)^2$
$(x+1)^2-(\sqrt{3}-1)^2=0$

$(x+1-\sqrt{3}+1)(x+1+\sqrt{3}-1)=0$

$(x+2-\sqrt{3})(x+\sqrt{3})=0$

$\Rightarrow x+2-\sqrt{3}=0$ hoặc $x+\sqrt{3}=0$

$\Rightarrow x=\sqrt{3}-2$ hoặc $x=-\sqrt{3}$

4 đúng không nhỉ?

\(\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)

Với \(x\ge3\Leftrightarrow\left|x-3\right|=x-3\)

Với \(x< 3\Leftrightarrow\left|x-3\right|=3-x\)

Ta có: \(\dfrac{\sqrt{a^2b}+b}{a-b}\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\div\dfrac{1}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{a\sqrt{b}+b}{a-b}\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\dfrac{a\sqrt{b}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\sqrt{\dfrac{\left(b-\sqrt{ab}\right)^2}{\left(a+\sqrt{b}\right)^2}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\dfrac{a\sqrt{b}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{ab}-b}{a+\sqrt{b}}\) vì \(a>b>0\)

\(=\dfrac{\left(a+\sqrt{b}\right)\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\sqrt{b}}{a+\sqrt{b}}\)

\(=b\)

Đpcm