\(\Leftrightarrow\frac{9\left(X+9\right)\left(X+9\right)\left(X+10\right)+10\left(X+10\right)\left(X+10\right)\left(X+9\right)}{90\left(X+10\right)\left(X+9\right)}=\frac{9.90\left(X+9\right)+10.90\left(X+10\right)}{90\left(X+10\right)\left(X+9\right)}\)

\(\Rightarrow9\left(X+9\right)^2\left(X+10\right)+10\left(X+10\right)^2\left(X+9\right)=810\left(X+9\right)+900\left(X+10\right)\)

\(\Leftrightarrow\left(9X+90\right)\left(X^2+18X+81\right)+\left(10X+90\right)\left(X^2+20X+100\right)=810X+7290+900X+9000\)

\(\Leftrightarrow\)9X³+162X²+729X+90X²+1620X+7290+10X³+200X²+1000X+90X²+1800X+9000=1710X+16290

\(\Leftrightarrow\)19X³+542X²+5149X+16290=1710X+16290

\(\Leftrightarrow\)19X³+542X²=16290-16290+1710X-5149X

\(\Leftrightarrow\)19X³+542X²=-3439X

\(\Leftrightarrow\)19X³+542X²+3439X=0

RỒI GIẢI TIẾP

nốt đi bạn

Mk nghĩ nên giải theo cách này thì hay hơn ( mk mớp 7 thui nên bài làm mang tính chất tham khảo nhé )

Ta có : 

\(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)

\(\Leftrightarrow\)\(\left(\frac{x+9}{10}+1\right)+\left(\frac{x+10}{9}+1\right)=\left(\frac{9}{x+10}+1\right)+\left(\frac{10}{x+9}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+19}{10}+\frac{x+19}{9}=\frac{x+19}{x+10}+\frac{x+19}{x+9}\)

\(\Leftrightarrow\)\(\frac{x+19}{10}+\frac{x+19}{9}-\frac{x+19}{x+10}-\frac{x+19}{x+9}=0\)

\(\Leftrightarrow\)\(\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\right)=0\)

Xét trường hợp \(x=0\)

\(\Rightarrow\)\(\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\right)=\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{10}-\frac{1}{9}\right)=\left(x+19\right).0=0\)

( NHẬN ) 

\(\Rightarrow\) Nếu \(x\ne0\) thì \(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\ne0\)

Xét trường hợp x nguyên dương ta có : 

\(\frac{1}{10}>\frac{1}{x+10}\)

\(\frac{1}{9}>\frac{1}{x+9}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}>0\)

Xét trường hợp x nguyên âm ta có : 

\(\frac{1}{10}< \frac{1}{x+10}\)

\(\frac{1}{9}< \frac{1}{x+9}\)

\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+9}-\frac{1}{x+10}< 0\)

Từ đó suy ra : 

\(x+19=0\)

\(\Rightarrow\)\(x=-19\)

Vậy \(x=0\) hoặc \(x=-19\)

a, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(1)

ĐKXĐ: \(\hept{\begin{cases}x+9\ne0\\x+10\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-9\\x\ne-10\end{cases}}}\)

(1)\(\Leftrightarrow\frac{9.\left(x+9\right)}{90}+\frac{10.\left(x+10\right)}{90}=\frac{9.\left(x+9\right)}{\left(x+9\right)\left(x+10\right)}+\frac{10.\left(x+10\right)}{\left(x+9\right)\left(x+10\right)}\)

\(\Leftrightarrow9.\left(x+9\right)+10.\left(x+10\right)=9.\left(x+9\right)+10.\left(x+10\right)\)

\(\Leftrightarrow9x+81+10x+100=9x+81+10x+100\)

\(\Leftrightarrow9x+10x-9x-10x=81+100-81-100\)

\(\Leftrightarrow0x=0\)

\(\Rightarrow x\in R\)trừ -9 và -10

\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)<=>  \(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

<=>\(\frac{8+x-8}{x-8}+\frac{11+x-11}{x-11}=\frac{9+x-9}{x-9}+\frac{10+x-10}{x-10}\)

<=>\(\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

<=>\(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=>\(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x-8}+\frac{1}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\end{cases}}\)

đến đoạn bạn giải tiếp nhé

Điều kiện:\(x\ne0\)

Đặt \(\frac{x}{3}-\frac{4}{x}=t\).Ta có:\(t^2=\left(\frac{x}{3}-\frac{4}{x}\right)^2=\frac{x^2}{9}-2.\frac{x}{3}.\frac{4}{x}+\frac{16}{x^2}=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\)

\(\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=t^2+\frac{8}{3}\).Thay vào pt ta có:\(t^2+\frac{8}{3}=\frac{10}{3}.t\)

\(\Leftrightarrow3t^2-10t+8=0\)\(\Leftrightarrow3t^2-4t-6t+8=0\)

\(\Leftrightarrow t\left(3t-4\right)-2\left(3t-4\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\Rightarrow\orbr{\begin{cases}t=2\\t=\frac{4}{3}\end{cases}}\)

Với \(t=2\) thì \(\frac{x^2-12}{3x}=2\Leftrightarrow x^2-12-6x=0\)\(\Rightarrow x^2-6x+9-21=0\)

\(\Leftrightarrow\left(x-3\right)^2=21\Rightarrow\orbr{\begin{cases}x-3=\sqrt{21}\\x-3=-\sqrt{21}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{21}+3\\x=3-\sqrt{21}\end{cases}}\)

Với \(t=\frac{4}{3}\) thì \(\frac{x^2-12}{3x}=\frac{4}{3}\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

Tập nghiệm của pt S=\(\left\{\sqrt{21}+3;3-\sqrt{21};-2;6\right\}\)

\(\frac{x^2}{9}+\frac{16}{x^2}=\frac{10}{3}\left(\frac{x}{3}-\frac{4}{x}\right)\)

\(\Leftrightarrow\frac{x^2}{9}-\frac{10x}{9}+\frac{40}{3x}+\frac{16}{x^2}=0\)

\(\Leftrightarrow\frac{x^4-10x^3+120x+144}{9x^2}=0\)

\(\Leftrightarrow x^4-10x^3+120x+144=0\)

\(\Leftrightarrow x^4-6x^3-12x^2-4x^3+24x^2+48x-12x^2+72x+144=0\)

\(\Leftrightarrow x^2\left(x^2-6x-12\right)-4x\left(x^2-6x-12\right)-12\left(x^2-6x-12\right)=0\)

\(\Leftrightarrow\left(x^2-4x-12\right)\left(x^2-6x-12\right)=0\)

\(\Leftrightarrow\left(x^2+2x-6x-12\right)\left(x^2-6x-12\right)=0\)

\(\Leftrightarrow\left[x\left(x+2\right)-6\left(x+2\right)\right]\left(x^2-6x-12\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+2\right)\left(x^2-6x-12\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-6=0\\x+2=0\\x^2-6x-12=0\left(1\right)\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-2\end{array}\right.\)(tm)

\(\Delta_{\left(1\right)}=\left(-6\right)^2-\left(-4\left(1.12\right)\right)=84\)

\(\Rightarrow\)\(x_{1,2}=\frac{6\pm\sqrt{84}}{2}\) (tm)

Vậy pt có nghiệm là \(x=-2;x=6\)và \(x=\frac{6\pm\sqrt{84}}{2}\)