ĐK: x>0

Đặt a=1/x ta được: a>0

\(a+\frac{1}{3}=\sqrt{\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}}\)

\(\Leftrightarrow a^2+\frac{1}{9}+\frac{2}{3}a=\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{2}{3}a=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a.\left(a+\frac{2}{3}\right)=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a+\frac{2}{3}=\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{4}{9}+\frac{4}{3}a=\frac{4}{9}+2a^2\)

<=>\(a^2-\frac{4}{3}a=0\Leftrightarrow a=0\left(loại\right);a=\frac{4}{3}\)

<=>\(x=\frac{3}{4}\)(loại -3/2)

Vậy x=3/4

= $\frac{3+x}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}$3+x3x =√19 ‍+1x √49 +2x2 

Thực ra cũng EZ thôi :

\(\frac{6}{x^2-9}-1+\frac{4}{x^2-11}-1-\frac{7}{x^2-8}+1-\frac{3}{x^2-12}+1=0=>\)

\(\frac{15-x^2}{x^2-9}+\frac{15-x^2}{x^2-11}-\frac{15-x^2}{x^2-8}-\frac{15-x^2}{x^2-12}=0\)

=> \(\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}-\frac{1}{x^2-8}-\frac{1}{x^2-12}\right)=0\)

=>\(15-x^2=0=>x=\pm\sqrt{15}\)

Hình như còn nghiệm , any body help me ?

\(x\ne\pm2\)

Đặt \(\left\{{}\begin{matrix}\frac{x+3}{x-2}=a\\\frac{x-3}{x+2}=b\end{matrix}\right.\) phương trình trở thành:

\(a^2+6b^2=7ab\)

\(\Leftrightarrow a^2-7ab+6b^2=0\)

\(\Leftrightarrow a^2-ab-6ab+6b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-6b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-6b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=6b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=\frac{6\left(x-3\right)}{x+2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}5x=-5x\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=6\end{matrix}\right.\)

ĐKXĐ: \(-3\le x\le3;x\ne0\)

Đặt \(\sqrt{9-x^2}=a\left(a\ge0;a\ne3\right)\Rightarrow x^2=9-a^2\),khi đó pt đã cho trở thành:

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

\(\Rightarrow3-a+\frac{1}{4\left(3-a\right)}=1\)

\(\Rightarrow\frac{4\cdot\left(3-a\right)^2+1}{4\left(3-a\right)}=1\Rightarrow4a^2-24a+37=12-4a\)

\(\Rightarrow4a^2-20a+25=0\Rightarrow\left(2a-5\right)^2=0\Rightarrow2a-5=0\)

\(\Rightarrow a=\frac{5}{2}\)(tm điều kiện),theo cách đặt ta có

\(\sqrt{9-x^2}=\frac{5}{2}\Rightarrow9-x^2=\frac{25}{4}\Rightarrow x^2=\frac{11}{4}\Rightarrow x=\frac{\sqrt{11}}{2}\)(TMĐKXĐ)

Vậy pt đã cho có nghiệm duy nhất là \(x=\frac{\sqrt{11}}{2}\)

Cái này bạn đặt x+3/x-2 = a 

x-3/x+2 = b

=> x^2-9/x^2-4 = ab

Ta có : a^2 - 7ab + 6b^2 = 0

<=> a^2 - 6ab - ab + 6b^2 = 0

PT đa thức thành nhân tử là xong :D 

ĐKXĐ: \(\left\{{}\begin{matrix}x^2\ne9\\x^2\ne11\\x^2\ne8\\x^2\ne12\end{matrix}\right.\Leftrightarrow x\notin\left\{3;-3;\sqrt{11};-\sqrt{11};2\sqrt{2};-2\sqrt{2};2\sqrt{3};-2\sqrt{3}\right\}\)

Đặt \(x^2-11=a\)(Điều kiện: \(a\notin\left\{-2;0;-3;1\right\}\))

PT\(\Leftrightarrow\frac{6}{a+2}+\frac{4}{a}-\frac{7}{a+3}-\frac{3}{a-1}=0\)

\(\Leftrightarrow\frac{6}{a+2}-1+\frac{4}{a}-1+\frac{-7}{a+3}+1+\frac{-3}{a-1}+1=0\)

\(\Leftrightarrow\frac{6-a-2}{a+2}+\frac{4-a}{a}+\frac{-7+a+3}{a+3}+\frac{-3+a-1}{a-1}=0\)

\(\Leftrightarrow-\frac{a-4}{a+2}-\frac{a-4}{a}+\frac{a-4}{a+3}+\frac{a-4}{a-1}=0\)

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

\(\Leftrightarrow a-4=0\)

hay a=4

\(\Leftrightarrow x^2-11=4\)

\(\Leftrightarrow x^2=15\)

hay \(x=\pm\sqrt{15}\)

Điều kiện: x \(\ne\) 2; -2

Đặt \(\frac{x+3}{x-2}=a;\frac{x-3}{x+2}=b\). Khi đó, PT trở thành: 3a² + 168b² - 46ab = 0 <=> 3a² - 46ab + 168b² = 0  (1)

Coi a là ẩn, b là tham số

\(\Delta\) = (-46b)² - 4.3.168b² = 100b² . (1) có 2 nghiệm là: 

a = \(\frac{46b+10b}{6}=\frac{28b}{3}\) hoặc a = \(\frac{46b-10b}{6}=6b\)

+) Nếu a = 6b thì \(\frac{x+3}{x-2}=6.\frac{x-3}{x+2}\) <=> (x+3)(x+2) = 6(x - 3)(x - 2) <=> x² + 5x + 6 = 6x² - 30x + 36

<=> 5x² - 35x + 30 = 0 <=> x² - 7x + 6 = 0 <=> x = 1 hoặc x = 6 (thỏa mãn)

+) nếu a = \(\frac{28}{3}\)b : Giải tương tự:....

Vậy.........

\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)

\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)

\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32

\(\Leftrightarrow\) 3x² - 6x + 2x - 4 + 2x² + 8x + x + 4 = 5x² - 10x + 20x - 40 - x + 32

\(\Leftrightarrow\) 5x² + 5x = 5x² + 9x - 8

\(\Leftrightarrow\) 5x² + 5x - 5x² - 9x + 8 = 0

\(\Leftrightarrow\) -4x + 8 = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)

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

\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)

\(\Leftrightarrow\) x² - 3x + 2mx - 6m + x² + 3x - mx - 3m - mx² - mx = 0

\(\Leftrightarrow\) (2 - m)x² - 9m = 0

Thay m = 1 ta được:

(2 - 1)x² - 9 . 1 = 0

\(\Leftrightarrow\) x² - 9 = 0

\(\Leftrightarrow\) (x - 3)(x + 3) = 0

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

Vậy S = \(\varnothing\)

Thay m = 2 ta được:

(2 - 2)x² - 9 . 2 = 0

\(\Leftrightarrow\) -18 = 0

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!!