\(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\frac{-9}{2}+\frac{1}{7}.\frac{-9}{10}\)

\(=\frac{-9}{10}.\frac{5}{14}+\frac{-9}{10}.\frac{1}{2}+\frac{1}{7}.\frac{-9}{10}\)

\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{2}+\frac{1}{7}\right)=\frac{-9}{10}.1=\frac{-9}{10}\)

Đề bài là gì bạn ơi có chỗ ...

chưa có đề bạn ơi! Y_Y

Hok tốt ^_^

Quy luật của dãy số này là gì bạn có thể nói rõ ra đc ko

\(=\frac{5}{9}.\frac{-3}{11}-\frac{5}{9}.\frac{10}{11}+\frac{5}{9}.\frac{10}{11}+\frac{5}{9}.\frac{2}{11}=\frac{5}{9}.\left(\frac{-3}{11}-\frac{10}{11}+\frac{10}{11}+\frac{2}{11}\right)=\frac{5}{9}.\left(\frac{-1}{11}\right)=\frac{-5}{99}\)

       \(\frac{1}{10}+\frac{9}{10}+\frac{10}{10}+9+\frac{160}{2}\)

\(=\frac{10}{10}+\frac{10}{10}+9+\frac{160}{2}\)

\(=1+1+9+80\)

\(=91\)

Ta có \(\frac{1}{10}+\frac{9}{10}+\frac{10}{10}+9+\frac{160}{2}=\left(\frac{1}{10}+\frac{9}{10}\right)+\left(\frac{10}{10}+9\right)+\frac{160}{2}\)

\(=1+10+80\)

\(=91\)

Vậy giá trị biểu thức là 91 

\(\frac{1}{10}+\frac{9}{10}+\frac{10}{10}+9+\frac{160}{2}\)

\(=\) \(2+9+80\)

\(=\)\(91\)

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

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

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

\(\Leftrightarrow\)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

= \(\frac{5\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}{10\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}\)

=\(\frac{5}{10}\)

=\(\frac{1}{2}\)