mình lười nên nói cách làm nhé

B1: chuyển \(\dfrac{6}{x^2-9}\)sang vế trái và thêm dấu trừ ở trc \(\dfrac{6}{x^2-9}\)và vế phải =0

B2: để ý thấy \(x^2-9\)=(x-3).(x+3) tức là hằng đẳng thức số 3 ý

B3: quy đồng mẫu , mẫu số chung là (x-3).(x+3).(2x+7)

B4: chia cả hai vế cho (x-3).(x+3).(2x+7)

lưu ý : bước này là dấu⇒ chứ ko phải dấu ⇔ nhé

B5: giải pt như bình thg thui

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

Ta có: \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\)

\(\Leftrightarrow\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

Suy ra: \(13x+39+x^2-9=12x+42\)

\(\Leftrightarrow x^2+13x+30-12x-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

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

Vậy: S={-4}

Đặt t=x²-2x+3(t\(\ge\)2)

PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)

<=>2t²+2t+2t²-2=9t²-9

<=>5t²-2t-7=0

<=>(t+1)(5t-7)=0

Do t\(\ge\)2

=>t+1>0 5t-7>0

Vậy pt vô nghiệm

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

Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)

Thì ta có:

\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)

\(\Leftrightarrow5t^2-t-4=0\)

\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)

\(\Rightarrow x^2-2x+2=1\)

\(\Leftrightarrow x=1\)

Vậy PT có 1 nghiệm là x = 1

* Nếu \(x< 2\) thì phương trình trở thành:

\(\Leftrightarrow\left(2-x\right)+\left(3-x\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow2-x+3-x+8-2x=9\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\) (nhận)

* Nếu \(2\le x\le3\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(3-x\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow x-2+3-x+8-2x=9\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\) (loại)

*Nếu \(3\le x\le4\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(x-3\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow x-2+x-3+8-2x=9\)

\(\Leftrightarrow3=9\) (vô lí)

* Nếu \(x>4\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(x-3\right)+\left(2x-8\right)=9\)

\(\Leftrightarrow x-2+x-3+2x-8=9\)

\(\Leftrightarrow4x=21\)

\(\Leftrightarrow x=\dfrac{21}{4}\)(nhận)

Vậy phương trình có tập nghiệm \(S=\left\{1;\dfrac{21}{4}\right\}\)

Làm cho bạn 1 con thôi dài quá trôi hết màn hình:

c) có vẻ khó nhất (con khác tương tự)

đặt 2x+2=t=> x+1=t/2

\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)

\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)

<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)

2.

ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)

\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)

\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)

\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)

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

\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)

\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)

\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

ĐKXĐ: ...

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

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

a, Ta có: \(\Delta'=1-m+3=4-m\)

Phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow4-m>0\Leftrightarrow m< 4\)

b, ĐXXĐ: \(x\le\frac{9}{4}\)

\(pt\Leftrightarrow\sqrt{\left(9-4x\right)\left(x-3\right)^2}=\left|-2x+5\right|\sqrt{9-4x}\)

\(\Leftrightarrow\sqrt{9-4x}\left(\left|x-3\right|-\left|-2x+5\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\\left|x-3\right|=\left|-2x+5\right|\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\x-3=-2x+5\\x-3=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{4}\\x=\frac{8}{3}\left(l\right)\\x=2\end{matrix}\right.\)

Vậy pt đã cho có 2 nghiệm \(x=2;x=\frac{9}{4}\)