ĐKXĐ: \(x\in R\)

\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)

=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)

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

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-x^3+2x^2-16}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-2x^3+x^3-2x^2+4x^2-8x+8x-16}{\sqrt{x^4-x^3+2x^2}+4}\)

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

=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)

=>x-2=0

=>x=2(nhận)

\(3x^2-5x+6=2x\sqrt{x^2-x+2}\)

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

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

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

Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

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

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

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

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

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

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

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

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

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

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

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

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

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

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

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

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

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

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

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

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

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

Bạn coi lại đề xem có sai không chứ nghiệm giải ra xấu cực. Và phương trình không rút gọn hết nghe cũng rất vô lý.

Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ

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

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

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

Cái ngoặc to yên tâm là vô nghiệm từ đó...

P/s: em chi có mỗi cách này thôi, ko biết có đúng không nữa..

Anh ko hiểu từ dòng đầu qua cái p/s ở dòng 2 luôn @@

a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)

\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)

\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)

\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)

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

\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)

\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)

\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)

b. ĐKXĐ: \(x\ge5\)

\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)

\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)

\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2+3b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)