Đặt \(x^2+1=a\)

\(\Rightarrow\frac{a}{120}+\frac{a+1}{119}+\frac{a+2}{118}=3\)

\(\Leftrightarrow21241a=2506200\)

\(\Leftrightarrow a=\frac{2506200}{21241}\)

\(\Rightarrow x=.....\)

\(\frac{x^2}{120}+\frac{x^2+1}{119}+\frac{x^2+2}{118}=3\)

\(\Leftrightarrow\frac{x^2}{120}+1+\frac{x^2+1}{119}+1+\frac{x^2+2}{118}+1=6\)

\(\Leftrightarrow\frac{x^2+120}{120}+\frac{x^2+120}{119}+\frac{x^2+120}{118}=6\)

\(\Leftrightarrow\left(x^2+120\right)\left(\frac{1}{120}+\frac{1}{119}+\frac{1}{118}\right)=6\)

\(\Leftrightarrow x^2+120=\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}\)

\(\Leftrightarrow x^2=\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}-1\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}-1}\\x=-\sqrt{\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}-1}}\end{cases}}\)

ĐK:.....

\(\left(\frac{1}{x}+\frac{1}{x+7}\right)+\left(\frac{1}{x+2}+\frac{1}{x+5}\right)=\left(\frac{1}{x+1}+\frac{1}{x+6}\right)+\left(\frac{1}{x+3}+\frac{1}{x+4}\right)\)

=> \(\frac{2x+7}{x\left(x+7\right)}+\frac{2x+7}{\left(x+2\right)\left(x+5\right)}=\frac{2x+7}{\left(x+1\right)\left(x+6\right)}+\frac{2x+7}{\left(x+3\right)\left(x+4\right)}\)

=> \(\left(2x+7\right)\left(\frac{1}{x\left(x+7\right)}+\frac{1}{\left(x+2\right)\left(x+5\right)}-\frac{1}{\left(x+1\right)\left(x+6\right)}-\frac{1}{\left(x+3\right)\left(x+4\right)}\right)=0\)

=> 2x + 7 = 0 hoặc \(\frac{1}{x\left(x+7\right)}+\frac{1}{\left(x+2\right)\left(x+5\right)}-\frac{1}{\left(x+1\right)\left(x+6\right)}-\frac{1}{\left(x+3\right)\left(x+4\right)}=0\)

+)  2x + 7 = 0 => x = -7/2 (T/m)

+) \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}=0\) (*)

Đặt t = x² + 7x . Khi đó pt có dạng

\(\frac{1}{t}+\frac{1}{t+10}-\frac{1}{t+6}-\frac{1}{t+12}=0\)

=> (t + 10)(t + 6)(t + 12) + t(t + 6)(t + 12) - t(t + 10)(t + 12) - t(t + 10)(t + 6) = 0 

=> [(t + 10)(t + 6)(t + 12) - t(t + 10)(t + 12)] + [t(t + 6)(t + 12) - t(t + 10)(t + 6)] = 0 

=> 6(t + 10)(t + 12) + 2t(t + 6) = 0 

<=> 6t² + 132t  + 720 + 2t² + 12t = 0 

=> 8t² + 144t + 720 = 0  (PT này vô nghiêm)

=> (*) Vô nghiệm

Vậy PT đã cho có nghiệm là x = -7/2

Điều kiện \(x\ne1.\)

Đặt \(y=\frac{x}{x-1}\to xy=x+y\) và \(x^3+y^3+3xy=2\) . Từ đây cho ta \(\left(x+y\right)^3-3xy\left(x+y\right)+3xy=2\to t^3-3t^2+3t=2\), với \(t=xy\), hay \(t^3-3t^2+3t-1=1\Leftrightarrow\left(t-1\right)^3=1\Leftrightarrow t-1=1\Leftrightarrow t=2.\)

Vậy ta được \(x+y=xy=2\to x\left(2-x\right)=2\to x^2-2x+2=0\) phương trình cuối vô nghiệm nên phương trình đã cho vô nghiệm

bằng 1 nữa nha

lớp 9 a

ĐK: \(x\ge0\)

\(PT\Leftrightarrow\frac{\sqrt{x+3}-\sqrt{x+2}}{1}+\frac{\sqrt{x+2}-\sqrt{x+1}}{1}+\frac{\sqrt{x+1}-\sqrt{x}}{1}=1\)

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

\(\Leftrightarrow x+3+x-2\sqrt{x^2+3x}=1\)\(\Leftrightarrow2x+2=2\sqrt{x^2+3x}\)

\(\Leftrightarrow x^2+2x+1=x^2+3x\)

\(\Leftrightarrow x=1\)

Vậy.........................

ĐK: \(x\ne1\)

\(pt\Leftrightarrow x^3\left(x-1\right)^3+x^3+3x^2\left(x-1\right)^2-2\left(x-1\right)^3=0\)

\(\Leftrightarrow\left(x^2-2x+2\right)\left(x^4-x^3+2x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[\left(x-1\right)^2+1\right]\left[\left(x^2-\frac{x}{2}\right)^2+\frac{3x^2}{4}+\left(x+1\right)^2\right]=0\)

\(\Leftrightarrow x^2-\frac{x}{2}=x=x+1=0\text{ (vô nghiệm)}\)

Vậy pt vô nghiệm.

b) ĐKXĐ:    \(x\ne1\)

Ta có:

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

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

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

Đặt \(\frac{x^2}{x-1}=a\)

Khi đó pt đã cho trở thành:

\(a^3-3a^2+3a-2=0\)

\(\Leftrightarrow\left(a-1\right)^3=1\Rightarrow a-1=1\Leftrightarrow a=2\)

Theo cách đặt:   \(\frac{x^2}{x-1}=2\Rightarrow x^2=2x-2\Leftrightarrow x^2-2x+1=-1\Leftrightarrow\left(x-1\right)^2=-1\left(ptvn\right)\)

a) ĐKXĐ:   \(x\ge8\)

Ta có:

\(x-\sqrt{x-8}-3\sqrt{x}+1=0\)

\(\Leftrightarrow x-9-\left(\sqrt{x-8}-1\right)-3\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow x-9-\frac{x-9}{\sqrt{x-8}+1}-3.\frac{x-9}{\sqrt{x}+3}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1=0\end{cases}}\)

+)  \(x-9=0\Leftrightarrow x=9\left(TMĐKXĐ\right)\)

+)  \(\frac{3}{\sqrt{x}+3}=\frac{\sqrt{x-8}}{\sqrt{x-8}+1}\Rightarrow\sqrt{x\left(x-8\right)}=3\)

\(\Leftrightarrow x^2-8x-9=0\Leftrightarrow\orbr{\begin{cases}x=9TMĐKXĐ\\x=-1\left(KTMĐKXĐ\right)\end{cases}}\)

Vaayh pt có 1 nghiệm là x=9

\(C2:\text{ }pt\Leftrightarrow x\left(3-x\right)\left[x\left(x+1\right)+3-x\right]=2\left(x+1\right)^2\)

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

\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+2\right)=0\)

\(\Leftrightarrow x=1\)

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

\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)

\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0\Rightarrow!2x+1!=2x+1\)

\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)

\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)

\(\left\{\begin{matrix}2x+1=0\\-x^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-\frac{1}{2}\\x=0\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

\(ĐKXĐ:x\ne\frac{5-\sqrt{13}}{2};x\ne\frac{5+\sqrt{13}}{2}\)

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

*) Xét x = 0 thì \(\frac{4x}{x^2+x+3}+\frac{5x}{x^2-5x+3}=0\)(Loại)

*) Xét \(x\ne0\)thì phương trình tương đương \(\frac{4}{x+\frac{3}{x}+1}+\frac{5}{x+\frac{3}{x}-5}=-\frac{3}{2}\)

Đặt \(x+\frac{3}{x}=t\)thì phương trình trở thành \(\frac{4}{t+1}+\frac{5}{t-5}=-\frac{3}{2}\)

\(\Leftrightarrow\frac{4t-20+5t+5}{\left(t+1\right)\left(t-5\right)}=-\frac{3}{2}\Leftrightarrow\frac{9t-15}{t^2-4t-5}=-\frac{3}{2}\)

\(\Leftrightarrow18t-30=-3t^2+12t+15\Leftrightarrow3t^2+6t-45=0\)

\(\Leftrightarrow3\left(t-3\right)\left(t+5\right)=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=-5\end{cases}}\)

+) t = 3 thì \(x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2-3x+3=0\)

Mà \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0\forall x\)nên loại trường hợp t = 3

+) t = -5 thì \(x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+5x+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{13}}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)

Vậy phương trình có 2 nghiệm \(\left\{\frac{-5+\sqrt{13}}{2};\frac{-5-\sqrt{13}}{2}\right\}\)

Bài làm:

đkxđ: \(x\ne\left\{\frac{5+\sqrt{13}}{2};\frac{5-\sqrt{13}}{2}\right\}\)

+ Nếu x = 0:

\(Pt\Leftrightarrow0=-\frac{3}{2}\)(vô nghiệm)

+ Nếu x khác 0:

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

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

Đặt \(x+\frac{3}{x}=y\)

\(Pt\Leftrightarrow\frac{4}{y+1}+\frac{5}{y-5}=-\frac{3}{2}\)

\(\Leftrightarrow\frac{8\left(y-5\right)+10\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}=-\frac{3\left(y-5\right)\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}\)

\(\Rightarrow8y-40+10y+10=-3\left(y^2-4y-5\right)\)

\(\Leftrightarrow18y-30=-3y^2+12y+15\)

\(\Leftrightarrow3y^2+6y-45=0\)

\(\Leftrightarrow y^2+2y-15=0\)

\(\Leftrightarrow\left(y-3\right)\left(y+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-3=0\\y+5=0\end{cases}}\Leftrightarrow\Leftrightarrow\orbr{\begin{cases}y=3\\y=-5\end{cases}}\)

Nếu: \(y=3\Leftrightarrow x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2+3=3x\)

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

\(\Leftrightarrow\left(x^2-3x+\frac{9}{4}\right)+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-\frac{3}{4}\)(vô lý)

=> không tồn tại x thỏa mãn

Nếu: \(y=-5\Leftrightarrow x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+3=-5x\)

\(\Leftrightarrow x^2+5x+3=0\)

\(\Leftrightarrow\left(x^2+5x+\frac{25}{4}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}-\frac{\sqrt{13}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{13}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5-\sqrt{13}}{2}=0\\x+\frac{5+\sqrt{13}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-5}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)(thỏa mãn)

Vậy tập nghiệm của PT \(S=\left\{\frac{-5-\sqrt{13}}{2};\frac{\sqrt{13}-5}{2}\right\}\)