ĐKXĐ: \(x\ge\dfrac{1}{3}\)

PT \(\Leftrightarrow2\left(x-\sqrt{3x-1}\right)+\left[\left(2x+1\right)-\sqrt{3x^2+7x}\right]=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+1\right)}{x+\sqrt{3x-1}}+\dfrac{\left(2x+1\right)^2-\left(3x^2+7x\right)}{2x+1+\sqrt{3x^2+7x}}=0\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left[\dfrac{2}{x+\sqrt{3x-1}}+\dfrac{1}{2x+1+\sqrt{3x^2+7x}}\right]=0\)

Cái ngoặc to vô nghiệm, đến đây bạn có thể giải.

3x2−7x+3−3x2−5x−1=x2−2−x2−3x+4" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">3x2−7x+3−3x2−5x−1=x2−2−x2−3x+4
⇔−2x+43x2−7x+3+3x2−5x−1=3x−6x2−2+x2−3x+4" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">⇔−2x+43x2−7x+3+3x2−5x−1=3x−6x2−2+x2−3x+4

⇔(x−2)(3x2−2+x2−3x+4+23x2−7x+3+3x2−5x−1)=0" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">⇔(x−2)(3x2−2+x2−3x+4+23x2−7x+3+3x2−5x−1)=0
Đến đây thì bạn có thể suy ra nghiệm của phương trình sau cùng là x=2. Kiểm tra lại điều kiện ban đầu để kết luận nghiệm của phương trình đã cho.
(2)đk:23≤x≤7

3x−183x−2+4+x−67−x−1+(x−6)(3x2+x−2)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">3x−183x−2+4+x−67−x−1+(x−6)(3x2+x−2)=0

⇔(x−6)(33x−2+4+17−x−1+3x2+x−2)" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">⇔(x−6)(33x−2+4+17−x−1+3x2+x−2)=0

⇔x=6" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">⇔x=6

23≤x≤7" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">23≤x≤7

(33x−2+4+17−x−1+3x2+x−2)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">(33x−2+4+17−x−1+3x2+x−2)>0

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

 

  

ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)

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

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

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

\(\Leftrightarrow x-5=0\) (do \(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\))

\(\Rightarrow x=5\)

ĐKXĐ: \(\left\{{}\begin{matrix}3x+1>=0\\6-x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\x< =6\end{matrix}\right.\)

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

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

=>\(\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+3x^2-15x+x-5=0\)

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

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

=>x-5=0

=>x=5(nhận)

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow5x-1=3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

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

\(\Leftrightarrow x^2+4x+4=4\left(3x-2\right)\left(x-1\right)\)

\(\Leftrightarrow11x^2-24x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{11}\left(loại\right)\\x=2\end{matrix}\right.\)