Link tham khảo: https://diendantoanhoc.net/topic/134563-gi%E1%BA%A3i-ph%C6%B0%C6%A1ng-tr%C3%ACnh-fracx-3x-23-x-3316/

ko bít làm bn ạ !

Lời giải:

Đặt \(x-2=a(a\neq 0)\). PT trở thành:

\(\left(\frac{a-1}{a}\right)^3-(a-1)^3=16\)

\(\Leftrightarrow 16+(a-1)^3-\left(1-\frac{1}{a}\right)^3=0\)

\(\Leftrightarrow 16+a^3-3a^2+3a-1-\left(1-\frac{3}{a}+\frac{3}{a^2}-\frac{1}{a^3}\right)=0\)

\(\Leftrightarrow a^3+\frac{1}{a^3}-3(a^2+\frac{1}{a^2})+3(a+\frac{1}{a})+14=0\)

\(\Leftrightarrow (a+\frac{1}{a})^3-3(a^2+\frac{1}{a^2})+14=0\)

\(\Leftrightarrow (a+\frac{1}{a})^3-3(a+\frac{1}{a})^2+20=0\)

Đặt \(a+\frac{1}{a}=t\Rightarrow t^3-3t^2+20=0\)

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

\(\Leftrightarrow (t+2)(t^2-5t+10)=0\)

Dễ thấy \(t^2-5t+10>0, \forall t\in\mathbb{R}\Rightarrow t+2=0\Leftrightarrow t=-2\)

Do đó: \(a+\frac{1}{a}=-2\Leftrightarrow \frac{(a+1)^2}{a}=0\Rightarrow a=-1\)

\(\Rightarrow x=2+a=1\)

Vậy $x=1$ là nghiệm của pt.

kho thay duoc t^2-5t+10>0. t€R.

\(\Leftrightarrow\dfrac{3}{x-2}>0\)

=>x-2>0

hay x>2

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

Cảm ơn mn ạ.

ban da lam duoc chua vay, giup minh voi!

ban hoc  nha thay phuc nhom 1 ak

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)

\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)

\(x=\)-\(\dfrac{27}{10}\)

\(x:\dfrac{2}{9}=\dfrac{9}{2}\)

\(x.\dfrac{9}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}:\dfrac{9}{2}\)

\(x=1\)

\(\dfrac{x}{9}=\dfrac{5}{3}\)

\(x.3=5.9\)

\(x.3=45\)

\(x=45:3=15\)

\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)

\(x:\dfrac{8}{125}=\dfrac{125}{8}\)

\(x.\dfrac{125}{8}=\dfrac{125}{8}\)

\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)