\(\Leftrightarrow\frac{9\left(X+9\right)\left(X+9\right)\left(X+10\right)+10\left(X+10\right)\left(X+10\right)\left(X+9\right)}{90\left(X+10\right)\left(X+9\right)}=\frac{9.90\left(X+9\right)+10.90\left(X+10\right)}{90\left(X+10\right)\left(X+9\right)}\)
\(\Rightarrow9\left(X+9\right)^2\left(X+10\right)+10\left(X+10\right)^2\left(X+9\right)=810\left(X+9\right)+900\left(X+10\right)\)
\(\Leftrightarrow\left(9X+90\right)\left(X^2+18X+81\right)+\left(10X+90\right)\left(X^2+20X+100\right)=810X+7290+900X+9000\)
\(\Leftrightarrow\)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
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 \(x=-19\)
\(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)
\(\Leftrightarrow\frac{x+9}{10}.90\left(x+9\right)\left(x+10\right)+\frac{x+10}{9}.90\left(x+9\right)\left(x+10\right)\)\(=\frac{9}{x+10}.90\left(x+9\right)\left(x+10\right)+\frac{10}{x+9}.90\left(x+9\right)\left(x+10\right)\)
\(\Leftrightarrow9\left(x+9\right)^2\left(x+10\right)+10\left(x+9\right)\left(x+10\right)^2=810\left(x+9\right)+900\left(x+10\right)\)
\(\Leftrightarrow19x^3+542x^2+5149x=1710x+16290\)
\(\Leftrightarrow19x^3+542x^2+5194x=1710x+16290-16290\)
\(\Leftrightarrow19x^3+542x^2+5149x=1710x\)
\(\Leftrightarrow19x^3+542x^2+5149x=1710x-1710x\)
\(\Leftrightarrow19x^3+542x^2+3439x=0\)
\(\Leftrightarrow19x^3+542x^2+3439x=x\left(19x^2+542+3439\right)\)
\(\Leftrightarrow x\left(x+19\right)\left(19x+181\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=-19\\x=-\frac{181}{19}\end{cases}}\)
