ai làm ơn làm phước tick cho mk lên 190 với

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

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

\(\Leftrightarrow2x^2+2-2x^2-2x=0\)

\(\Leftrightarrow-2x+2=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)