\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)

Ta có : 

\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)

\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)

\(=\)\(\frac{2}{7}-\frac{2}{7}\)

\(=\)\(0\)

Chúc bạn học tốt ~ 

thank nha

k cho mk nhé .OK

dài thế ai mà tính đc

\(\frac{1}{3}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4.(\frac{1}{9}-\frac{1}{7}-\frac{1}{11})}+\frac{3.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}{4.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}\)

\(A=\frac{1}{4}+\frac{3}{4}\)(Vì\(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\ne0\)và\(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\ne0\))

\(A=1\)

Vậy A = 1

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)

\(B=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-B=-\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{2.1}\)

\(-B=-\frac{1}{9}-\frac{1}{10}+\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+...+1-\frac{1}{2}\)

\(-B=-\frac{1}{9}.2-\frac{1}{10}+1\)

\(-B=-\frac{2}{9}-\frac{1}{10}+1\)

\(-B=\frac{-20}{90}-\frac{9}{90}+\frac{90}{90}\)

\(-B=\frac{61}{90}\)

\(B=\frac{-61}{90}\)

Vậy\(B=\frac{-61}{90}\)

Linz

=11/9+7/6*5/3

=11/9+7/6

=139/54

tự rút gọn nếu muốn nhé

k mình

=19/6 nha

HT

chúc bn năm mới vui vẻ

@@@@@@@@@@@@@@@@@@@

\(x\)là dấu nhân nha 

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\)

\(=\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+2\right)^2-\left(x-2\right)^2-16}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{8x-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8}{x+2}\)

bạn ơi, cho mình hỏi

4-x² thì sẽ = 2²-x² = (2-x)(2+x)

vậy sao bạn ghi là (x-2)(x+2) vậy???

Biến đổi như thế này nha bạn :

\(\frac{16}{4-x^2}=\frac{-1\cdot16}{-1\cdot\left(4-x^2\right)}\)

\(=\frac{-16}{-4+x^2}=\frac{-16}{x^2-4}\)

\(=\frac{-16}{\left(x-2\right)\left(x+2\right)}\)