a) We have \(3\le x\le5\)

Suppose \(\sqrt{5-x}=a\left(a\ge0\right);\sqrt{x-3}=b\left(b\ge0\right)\)

We imidiately have \(a+b=\sqrt{2}\)

On the other hand, we have \(a^2-b^2=\left(\sqrt{5-x}\right)^2-\left(\sqrt{x-3}\right)^2=\left(5-x\right)-\left(x-3\right)=8\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=8\) \(\Leftrightarrow\sqrt{2}\left(a-b\right)=8\) \(\Leftrightarrow a-b=4\sqrt{2}\)

And now we simply have \(\left\{{}\begin{matrix}a+b=\sqrt{2}\\a-b=4\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\sqrt{2}}{2}\left(take\right)\\b=\dfrac{3\sqrt{2}}{2}\left(take\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5-x}=\dfrac{5\sqrt{2}}{2}\\\sqrt{x-3}=\dfrac{3\sqrt{2}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-x=\dfrac{25}{2}\\x-3=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{2}\left(eliminate\right)\\x=\dfrac{15}{2}\left(take\right)\end{matrix}\right.\)

Therefore, this equation have the root \(x=\dfrac{15}{2}\)

b) We have \(x\ge2\)

\(\sqrt{x^2-4}=2\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-2\sqrt{x-2}=0\) \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\left(take\right)\)

So this equation has the root \(x=2\)

c) We have \(x\ge-\dfrac{4}{3}\). Suppose \(2x+3=A\left(A\ge0\right);3x+4=B\left(B\ge0\right)\). Notice that \(A+B=2x+3+3x+4=5x+7\). Thus, we can rewrite the equation as below:

\(\sqrt{A+B}=\sqrt{A}+\sqrt{B}\) (1)

Remember that \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\left(A,B\ge0\right)\). We can prove it easily by squaring each side of this inequality:

\(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2\ge\left(\sqrt{A+B}\right)^2\)\(\Leftrightarrow A+B+2\sqrt{AB}\ge A+B\Leftrightarrow2\sqrt{AB}\ge0\Leftrightarrow\sqrt{AB}\ge0\). This is always true when \(A,B\ge0\). "=" happens when one of the expression A, B is equal to 0.

Therefore, (1) happens when \(\left[{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(eliminate\right)\\x=-\dfrac{4}{3}\left(take\right)\end{matrix}\right.\)

Thus, this equation has the root \(x=-\dfrac{4}{3}\)

d) We have \(x\ge-\dfrac{1}{3}\)

We can see that \(x=-\dfrac{1}{11}\). So, \(\sqrt{5x+7}=\dfrac{6\sqrt{22}}{11}\); o\(\sqrt{x+3}=\dfrac{4\sqrt{22}}{11}\) and \(\sqrt{3x+1}=\dfrac{2\sqrt{22}}{11}\). Therefore, we can rewrite the equation as below:

\(\left(\sqrt{5x+7}-\dfrac{6\sqrt{22}}{11}\right)-\left(\sqrt{x+3}-\dfrac{4\sqrt{22}}{11}\right)-\left(\sqrt{3x+1}-\dfrac{2\sqrt{22}}{11}\right)=0\)Now you multiply and devide each of the terms by the conjugate expression. The first term will has \(5x+\dfrac{5}{11}=5\left(x+\dfrac{1}{11}\right)\) as the numerator, the second term's numerator will be \(x+\dfrac{1}{11}\), and the final term has \(3x+\dfrac{3}{11}=3\left(x+\dfrac{1}{11}\right)\) as the numerator. And now you can see the commom factor \(x+\dfrac{1}{11}\)

\(\left(\sqrt{A}-B=\dfrac{\left(\sqrt{A}-B\right)\left(\sqrt{A}+B\right)}{\sqrt{A}+B}=\dfrac{A-B^2}{\sqrt{A}-B}\right)\)

x³ + (x + 1)³ + (x + 2)³ 

= x³ + (x + 1)³ + (x + 2)³ - 3x(x + 1)(x + 2) + 3x(x + 1)(x + 2)

= \(\dfrac{\left(x+x+1+x+2\right)\left[\left(x-x-1\right)^2+\left(x-x-2\right)^2+\left(x+1-x-2\right)^2\right]}{2}+3x\left(x+1\right)\left(x+2\right)\)

\(=9\left(x+1\right)+3x\left(x+1\right)\left(x+2\right)⋮9\) (vì x(x + 1)(x + 2) tích 3 số nguyên liên tiếp) 

\(a,ĐKXĐ:x\ge\dfrac{3}{2}\)

\(\sqrt{2x-3}=2\sqrt{x}-2\)

\(2x-3=4x-8\sqrt{x}+4\)

\(2x-8\sqrt{x}+7=0\)

\(\sqrt{\Delta}=\sqrt{\left(-8\right)^2-4.2.7}=2\sqrt{2}\)

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+\sqrt{2}}{2}\left(TM\right)\)

\(x_2=\dfrac{4-\sqrt{2}}{2}\left(KTM\right)\)

\(b,\sqrt{\dfrac{4x+3}{\sqrt{x+1}}}=3\)

\(ĐKXĐ:x\ge-\dfrac{3}{4}\)

\(\dfrac{4x+3}{\sqrt{x+1}}=9\)

\(\left(4x+3\right)^2=81x+81\)

\(16x^2+24x+9=81x+81\)

\(16x^2-57x-72=0\)

\(\sqrt{\Delta}=9\sqrt{97}\)

\(x_1=\dfrac{57+9\sqrt{97}}{32}\left(TM\right)\)

\(x_2=\dfrac{57-9\sqrt{97}}{32}\left(KTM\right)\)

\(c,ĐKXĐ:x>1\)

\(\dfrac{\sqrt{x^2+3x-x-3}}{\sqrt{x-1}}=x+3\)

\(\dfrac{\sqrt{\left(x-1\right)\left(x+3\right)}}{\sqrt{x-1}}=x+3\)

\(\sqrt{x+3}=x+3\)

\(x+3=x^2+6x+9\)

\(x^2+5x+6=0\)

\(\left(x+3\right)\left(x+2\right)=0\)

\(\left[{}\begin{matrix}x=-3\left(KTM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

\(d,x>3\)

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

\(\dfrac{\sqrt{x-3}\sqrt{x-1}}{\sqrt{x-3}}=x-1\)

\(\sqrt{x-1}=x-1\)

\(\sqrt{x-1}=1\)

\(x=2\left(KTM\right)\)

\(\)

ĐKXĐ : \(x\le28\)

Đặt \(\sqrt[3]{x+5}=a;\sqrt{28-x}=b\left(b\ge0;a\le7\right)\)

Được a + b = 7 <=> b = 7 - a 

Lại có a³ + b² = 33

<=> a³ + (7 - a)² = 33

<=> a³ + a² - 14a + 16 = 0

<=> a³ - 8 + a² - 2a - 12a + 24 = 0

<=> (a - 2)(a² + 3a - 8) = 0

<=> \(\left[{}\begin{matrix}a=2\\a^2+3a-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=2\\a=\dfrac{\pm\sqrt{41}-3}{2}\end{matrix}\right.\)

Khi a = \(\dfrac{\sqrt{41}-3}{2}\Leftrightarrow b=\dfrac{17-\sqrt{41}}{2}\)

<=> \(x=28-\left(\dfrac{17-\sqrt{41}}{2}\right)^2=\dfrac{17\sqrt{41}-109}{2}\)(tm)

Khi a = \(\dfrac{-\sqrt{41}-3}{2}\Leftrightarrow b=\dfrac{17+\sqrt{41}}{2}\)

\(x=28-\left(\dfrac{17+\sqrt{41}}{2}\right)^2=\dfrac{-17\sqrt{41}-109}{2}\) (tm) 

a = 2 => x = 3 (tm)

Vậy tập nghiệm phương trình \(S=\left\{\dfrac{-17\sqrt{41}-109}{2};\dfrac{17\sqrt{41}-109}{2};3\right\}\)

a, \(\dfrac{\sqrt{x+1}}{\sqrt{x-1}}\) = 2 đk  x >1

\(\sqrt{\dfrac{x+1}{x-1}}\) = 2

⇔ \(\dfrac{x+1}{x-1}\) = 4

⇔ x + 1 = 4x - 4

3x = 5

x =5/3

b, \(\sqrt{\dfrac{x-1}{x+1}}\) = 2    đk  x ϵ { x ϵ R|  x≤-1. x >1}

⇔ \(\dfrac{x-1}{x+1}\) = 4

⇔  x - 1 = 4x + 4

⇔ 3x = -5

x = -5/3 

B = (sin²a + cos²a)² = 1² = 1

\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)

\(=\left(\sin^2\alpha+\cos\alpha^2\right)^2=1^2=1\)

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 ?