4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

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

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

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

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

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)

\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)

=>PTVN

b: 

\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)

=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)

\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)

PT có hai nghiệm là;

\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)

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

+) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

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

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

+) \(ĐKXĐ:\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)

\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)

\(M=\dfrac{a+2\sqrt{a}+a-2\sqrt{a}}{a-4}\cdot\dfrac{a-4}{2\sqrt{a}}\)

\(M=\dfrac{2a}{2\sqrt{a}}=\sqrt{a}\)

+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)

\(N=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(N=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4-x+9+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(N=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)

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

+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne4\end{matrix}\right.\)

 \(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)

\(Q=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)

\(Q=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4+\sqrt{x}-8-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(Q=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)

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

p/s: sorry tại n' câu wa nên mình ko làm chi tiết đc =(( lần sau nhớ chia các câu ra cho dễ nhìn hơn nha, đánh hơi mỏi tay :'( có j ko hỉu cmt dưới nha

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

cảm ơn bạn nha 

a) ĐK: \(x\ge3\)

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

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

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

     \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)

  Vậy ...

1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)

=>4x+8=3x-1

=>x=-9

2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)

=>8x-4=5x-7

=>3x=-3

=>x=-1

3: ĐKXD: x>=0

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

=>\(x+\sqrt{x}-6=x-1\)

=>căn x=-1+6=5

=>x=25

4: ĐKXĐ: x>=0

PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

=>x-2*căn x-3=x-4

=>-2căn x-3=-4

=>2căn x+3=4

=>2căn x=1

=>căn x=1/2

=>x=1/4

\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(=-\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)

\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\) (ĐK: \(x\ge0\))

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

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

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

______________

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

Đặt \(\sqrt{x-1}=a\), khi đó ta có:

\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

\(=\left[\dfrac{\sqrt{x-1}}{\sqrt{x-1}+3}+\dfrac{\left(x-1\right)+9}{9-\left(x-1\right)}\right]:\left[\dfrac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\dfrac{1}{\sqrt{x-1}}\right]\)

\(=\left(\dfrac{a}{a+3}+\dfrac{a^2+9}{9-a^2}\right):\left(\dfrac{3a+1}{a^2-3a}-\dfrac{1}{a}\right)\)

\(=\dfrac{a\left(3-a\right)+\left(a^2+9\right)}{\left(3+a\right)\left(3-a\right)}:\dfrac{\left(3a-1\right)-\left(a-3\right)}{a\left(a-3\right)}\)

\(=\dfrac{3a-a^2+a^2+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{3a-1-a+3}{a\left(a-3\right)}\)

\(=\dfrac{3a+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{2a+4}{a\left(a-3\right)}\)

\(=\dfrac{3\left(a+3\right)}{\left(a+3\right)\left(a-3\right)}.\dfrac{a\left(a-3\right)}{2\left(a+2\right)}\)

\(=\dfrac{-3a}{2\left(a+2\right)}\).

Suy ra: P \(=\dfrac{-3\sqrt{x-1}}{2\left(\sqrt{x-1}+2\right)}\).

Ta lại có: \(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)

\(=\sqrt[4]{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)^2}}-\sqrt[4]{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)^2}}\)

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

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

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

\(=2\).

Suy ra: \(P=\dfrac{-3\sqrt{2-1}}{2\left(\sqrt{2-1}+2\right)}=\dfrac{-3}{2.3}=-\dfrac{1}{2}\).

2. \(\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}\) (2)

\(\Leftrightarrow\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}-\dfrac{7}{\sqrt{x-3}}=0\)

\(\Leftrightarrow\dfrac{\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7}{\sqrt{x-3}}=0\)

\(\Leftrightarrow\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7=0\)

\(\Leftrightarrow\left|x\right|-16+\sqrt{x^2-9}-7=0\)

\(\Leftrightarrow\left|x\right|-23+\sqrt{x^2-9}=0\)

\(\Leftrightarrow\sqrt{x^2-9}=-\left|x\right|+23\)

\(\Leftrightarrow x^2-9=-\left(-\left|x\right|+23\right)^2\)

\(\Leftrightarrow x^2-9=-\left(-\left|x\right|\right)^2-46\cdot\left|x\right|+529\)

\(\Leftrightarrow x^2-9=\left|x\right|^2-46+\left|x\right|+529\)

\(\Leftrightarrow x^2-9=x^2-46\cdot\left|x\right|+529\)

\(\Leftrightarrow-9=-46\cdot\left|x\right|+529\)

\(\Leftrightarrow46\cdot\left|x\right|=529+9\)

\(\Leftrightarrow49\cdot\left|x\right|=538\)

\(\Leftrightarrow\left|x\right|=\dfrac{269}{23}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{269}{23}\\x=-\dfrac{269}{23}\end{matrix}\right.\)

Sau khi dùng phép thử ta nhận thấy \(x\ne-\dfrac{269}{23}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{269}{23}\right\}\)

3. sửa đề: \(\sqrt{14-x}=\sqrt{x-4}\sqrt{x-1}\) (3)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{\left(x-4\right)\left(x-1\right)}\)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-x-4x+4}\)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-5x+4}\)

\(\Leftrightarrow14-x=x^2-5x+4\)

\(\Leftrightarrow14-x-x^2+5x-4=0\)

\(\Leftrightarrow10+4x-x^2=0\)

\(\Leftrightarrow-x^2+4x+10=0\)

\(\Leftrightarrow x^2-4x-10=0\)

\(\Leftrightarrow x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot\left(-10\right)}}{2\cdot1}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{16+40}}{2}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{56}}{2}\)

\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{14}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{14}}{2}\\x=\dfrac{4+2\sqrt{14}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{14}\\x=2-\sqrt{14}\end{matrix}\right.\)

sau khi dùng phép thử ta nhận thấy \(x\ne2-\sqrt{14}\)

Vậy tập nghiệm phương trình (3) là \(S=\left\{2+\sqrt{14}\right\}\)