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

B=\(\frac{5}{9}\)x{\(\frac{7}{13}\)+\(\frac{9}{13}\)-\(\frac{3}{13}\)}

B=\(\frac{5}{9}\)x 1

B=\(\frac{5}{9}\)

Vậy B=\(\frac{5}{9}\)

Chúc bạn hok tốt

bằng 5/9 , chúc bạn học tốt !

B = \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

Ta thấy có phân số \(\frac{5}{9}\) là thừa số chung

\(\Leftrightarrow\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\left(\frac{7+9+3}{13}\right)\)

\(=\frac{5}{9}.\frac{19}{13}\left(\frac{5.19}{9.13}\right)\)

\(=\frac{95}{117}\)

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

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

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

\(\Leftrightarrow\)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 \(x=-19\)

ĐKXĐ: x≠-10; x≠-9

Ta có: \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{10}{x+9}+\frac{9}{x+10}\)

Vậy: x=0

Những phương trình là phương trình chính tắc của (H) là: b), c), d).

Từ \(\frac{9-x}{7}+\frac{11-x}{9}=2\)

\(=>\frac{9-x}{7}+\frac{11-x}{9}-2=0\)

\(=>\frac{9-x}{7}+\frac{11-x}{9}-1-1=0\)

\(=>\left(\frac{9-x}{7}-1\right)+\left(\frac{11-x}{9}-1\right)=0\)

\(=>\frac{2-x}{7}+\frac{2-x}{9}=0=>\left(2-x\right).\left(\frac{1}{7}+\frac{1}{9}\right)=0\)

Vì \(\frac{1}{7}+\frac{1}{9}\) khác 0=>2-x=0=>x=2

Theo T/c dãy tỉ số=nhau:

\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}\)\(=\frac{\left(x+y+z\right)+\left(16-25+9\right)}{9+16+25}=\frac{x+y+z}{50}\)

Thay x=2 vào \(\frac{x+16}{9}=>\frac{2+16}{9}=\frac{x+y+z}{50}=>\frac{x+y+z}{50}=2=>x+y+z=100\)

Vậy x+y+z=100

Ta có \(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

=> \(\frac{9}{25}x\left(1+\frac{3}{5}+1\right)+18\left(\frac{3}{5}+1\right)=x\)

=> \(\frac{117}{125}x+28,8=x\)

=> \(x-\frac{117}{125}x=28,8\)

=> \(\frac{8}{125}x=28,8\)

=> x = 450

Vậy x = 450

\(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

\(x\left(\frac{9}{25}+\frac{9}{25}+\frac{9}{25}\right).\frac{3}{5}+\frac{3}{5}.18+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+18\right)+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+\frac{450}{25}\right)+18=x\)

\(x.\frac{3}{5}.\frac{477}{25}+18=x\)

\(x.\frac{1431}{125}+\frac{2250}{125}=x\)

\(x.\frac{3681}{125}=x\)

vậy chac tui làm sai rồi