Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\)  = B

   Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)

                   \(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)

                     \(=4\)

     Mà \(B=2\Leftrightarrow A=2\)

Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$

$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$

$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$

$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$

$=>M=2$

\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

      \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

Bài 2:

\(B=\sqrt{28-16\sqrt{3}}+\sqrt{13-4\sqrt{3}}\)

\(=\sqrt{\left(4-2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-1\right)^2}\)

\(=\left|4-2\sqrt{3}\right|+\left|2\sqrt{3}-1\right|\)

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

\(=3\)

\(C=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\sqrt{2}.\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=5-3=2\)

\(D=\sqrt{4+2\sqrt{3}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)

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

\(=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

1) \(\Leftrightarrow x^2-7x+8+\sqrt{x^2-7x+8}-20=0\)

Đặt \(t=\sqrt{x^2-7x+8}\ge0\)

Phương trình tương đương

\(t^2+t-20=0\)

\(\left[{}\begin{matrix}t=4\left(TM\right)\\t=-5\left(KTM\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-7x+8}=4\)

Bạn đọc tự giải quyết tiếp bài toán.

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

\(\Leftrightarrow\dfrac{1}{x}+1=\sqrt[3]{2}\)

\(\Leftrightarrow\left(x+1\right)^3=2x^3\Leftrightarrow x^3-3x^2-3x-1=0\).

Do đó \(M=\dfrac{\sqrt{x^3+x^2+5x+3}-6}{\sqrt{x^3-2x^2-7x+3}}\)

\(M=\dfrac{\sqrt{\left(x^3-3x^2-3x-1\right)+\left(4x^2+8x+4\right)}-6}{\sqrt{\left(x^3-3x^2-3x-1\right)+\left(x^2-4x+4\right)}}\)

\(M=\dfrac{\sqrt{\left(2x+2\right)^2}-6}{\sqrt{\left(x-2\right)^2}}=\dfrac{2x+2-6}{x-2}=2\). (Do \(x>2\))

ta có:\(\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{1}{2}\)

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

\(\Rightarrow\hept{\begin{cases}x+1=3\sqrt{x}\\x-3\sqrt{x}=-1\end{cases}}\)

lại có \(B=\frac{3x\sqrt{x}+10x+19}{x^2+7x+15}\)

\(=\frac{3x\sqrt{x}-9x+19x+19}{x^2-9x+16x+15}\)

\(=\frac{3\sqrt{x}\left(x-3\sqrt{x}\right)+19\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(x-3\sqrt{x}\right)+16x+15}\)

\(=\frac{-3\sqrt{x}+19\times3\sqrt{x}}{-1\times\left(x+3\sqrt{x}\right)+16x+15}\)

\(=\frac{57\sqrt{x}-3\sqrt{x}}{15x+15-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{15\left(x+1\right)-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{45\sqrt{x}-3\sqrt{x}}\)

\(=\frac{54\sqrt{x}}{42\sqrt{x}}=\frac{27}{21}\)

Tham khảo:

Giải phương trình sau trên tập số thực: \(\frac{3(x^2+2x-3)}{\sqrt{x+4}-1}-\frac{7x^2-19x+12}{\sqrt{12-7x}}=16x^2+11x-27\) - ngọc trang

1)ĐK:`4x^2-12x+9>0`

`<=>(2n-3)^2>0`

`<=>2n-3 ne 0`

`<=>n ne 3/2`

`d)x^2-x+1`

`=(x-1/2)^2+3/4>0AAx`

`=>` bt xd `AAx in RR`

e)ĐK:`x^2-8x+15>0`

`<=>x^2-3x-5x+15>0`

`<=>x(x-3)-5(x-3)>0`

`<=>(x-3)(x-5)>0`

`TH1:` \(\begin{cases}x-3>0\\x-5>0\\\end{cases}\)

`<=>` \(\begin{cases}x>3\\x>5\\\end{cases}\)

`<=>x>5`

`TH2:` \(\begin{cases}x-3<0\\x-5<0\\\end{cases}\)

`<=>` \(\begin{cases}x<3\\x<5\\\end{cases}\)

`<=>x<3`

f)ĐK:`3x^2-7x+20>0`

`<=>x^2-2x+1+2x^2-5x+19>0`

`<=>(x-1)^2+2(x-5/2)^2+13/2>0` luôn đúng

c) Để biểu thức \(\dfrac{1}{\sqrt{4x^2-12x+9}}\) có nghĩa thì \(4x^2-12x+9>0\)

\(\Leftrightarrow\left(2x-3\right)^2>0\)

\(\Leftrightarrow2x-3\ne0\)

\(\Leftrightarrow2x\ne3\)

hay \(x\ne\dfrac{3}{2}\)

d) Để biểu thức \(\dfrac{1}{\sqrt{x^2-x+1}}\) có nghĩa thì \(x^2-x+1>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}>0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)(luôn đúng)

e) Để biểu thức \(\dfrac{1}{\sqrt{x^2-8x+15}}\) có nghĩa thì \(x^2-8x+15>0\)

\(\Leftrightarrow\left(x-4\right)^2>1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4>1\\x-4< -1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)

f) Để biểu thức \(\dfrac{1}{\sqrt{3x^2-7x+20}}\) có nghĩa thì \(3x^2-7x+20>0\)

\(\Leftrightarrow x^2-\dfrac{7}{3}x+\dfrac{20}{3}>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{191}{36}>0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{36}>0\)(luôn đúng)