1/ \(\sqrt{x-2}-\sqrt{1-3x}=0\\ đk:\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

=> pt vô no

2/ \(\sqrt{15-x}+\sqrt{3-x}=6\\ đk\left\{{}\begin{matrix}15-x\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le15\\x\le3\end{matrix}\right.\Leftrightarrow x\le3\)

\(pt\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow2\sqrt{\left(15-x\right)\left(3-x\right)}=2x+36\)

\(\Leftrightarrow4\left(15-x\right)\left(3-x\right)=\left(2x+18\right)^2\left(đk:x\ge-9\right)\)

\(\Leftrightarrow-144x=144\Leftrightarrow x=-1\left(nhan\right)\)

Câu 1: ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ \(\Rightarrow\) pt vô nghiệm

Câu 2:

ĐKXĐ: \(x\le3\)

\(\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow x+9=\sqrt{x^2-18x+45}\) (\(x\ge-9\))

\(\Leftrightarrow x^2+18x+81=x^2-18x+45\)

\(\Leftrightarrow36x=-36\Rightarrow x=-1\)

Câu 3:

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow x-1=4+x+1+4\sqrt{x+1}\)

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

Phương trình vô nghiệm

5.

ĐKXĐ: ...

\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)

\(\Leftrightarrow x=5\)

6.

ĐKXĐ: \(-4\le x\le4\)

\(\Leftrightarrow\frac{\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)

\(\Leftrightarrow\frac{x\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)

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

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

\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)

\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)

\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)

\(\Rightarrow x=\frac{96}{25}\)

1.

Bạn coi lại đề

2.

ĐKXĐ: \(1\le x\le2\)

Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:

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

\(\Leftrightarrow3\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)

\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)

\(\Leftrightarrow3\sqrt{2-x}+2-\sqrt{x+2}+1-\sqrt{x-1}=0\)

\(\Leftrightarrow3\sqrt{2-x}+\frac{2-x}{2+\sqrt{x+2}}+\frac{2-x}{1+\sqrt{x-1}}=0\)

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

\(\Leftrightarrow\sqrt{2-x}=0\Rightarrow x=2\)

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

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

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

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

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

Tự nhiên trả lời làm cái gì

Đăng lên để hỏi

Chứ không phải trả lời nha o0o I am a studious person CTV 

chuẩn không cần phải chỉnh nha bn!!!!

a)X=2,81376107

b)X=2

Bình phương đi bạn

TL:

1đk:x<1

.\(1+3x-1=9x^2\) 

     \(3x=9x^2\) 

   x=3x\(^2\) 

 =>x=0(ktm)  hoặc  x= \(\frac{1}{3}\left(tm\right)\) 

vậy x=\(\frac{1}{3}\) 

hc tốt:)

ĐKXĐ:....

a/ \(\sqrt{2x+3}=1+\sqrt{2}\Leftrightarrow2x+3=3+2\sqrt{2}\Rightarrow x=\sqrt{2}\)

b/ \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)

\(\Rightarrow\sqrt{3x}=4\sqrt{6}\Rightarrow3x=96\Rightarrow x=32\)

c/ \(\sqrt{3x-2}=2-\sqrt{3}\Rightarrow3x-2=7-4\sqrt{3}\)

\(\Rightarrow3x=9-4\sqrt{3}\Rightarrow x=\frac{9-4\sqrt{3}}{3}\)

d/ \(\sqrt{x-1}=\sqrt{5}-3\)

Do \(\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{5}-3< 0\end{matrix}\right.\) \(\Rightarrow\) phương trình vô nghiệm

1)\(\sqrt{2x^2-2x+\frac{1}{2}}=\frac{1}{\sqrt{2}}\left(ĐKXĐ:x^2-x+\frac{1}{4}\ge0\right)\)

   \(2x^2-2x+\frac{1}{2}=\frac{1}{2}\)

   \(2x^2-2x=0\)

    \(2x\left(x-1\right)=0\)

            \(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

2)\(\sqrt{9x-9}-2\sqrt{\frac{x-1}{4}}=6\left(ĐKXĐ:x\ge1\right)\)

    \(\sqrt{9\left(x-1\right)}-2.\frac{\sqrt{x-1}}{2}=6\)

   \(3\sqrt{x-1}-\left(\sqrt{x-1}\right)=6\)

  \(2\sqrt{x-1}=6\)

   \(\sqrt{x-1}=3=\sqrt{9}\)

    \(\Rightarrow x=10\)

4)\(1-3x+\sqrt{x^2-6x+9}=0\)

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

    \(1-3x+x-3=0\)

    \(x=-1\)

5)\(\frac{1}{2}\sqrt{\frac{3x+9}{4}}+\sqrt{x+3}=\sqrt{1-x}\)

    \(\frac{1}{2}.\frac{\sqrt{3x+9}}{2}+\sqrt{x+3}=\sqrt{1-x}\)

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

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

     \(\Rightarrow\sqrt{3}.\sqrt{x+3}+4\sqrt{x+3}=4\sqrt{1-x}\)

     \(\Rightarrow\left(\sqrt{3}+4\right)\left(\sqrt{x+3}\right)=\sqrt{2-2x}\)

6)\(\sqrt{4x^2-9}.\left(\sqrt{x+1}+1\right)=0\)

    \(\Rightarrow\orbr{\begin{cases}4x^2-9=0\\\sqrt{x+1}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x^2=9\\\sqrt{x+1}=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)