thấy B/A

 lớn hơn 1 .Vây  B

 lớn hơn A

\(A=\frac{10^9+5}{10^9-2}\)                                                                                                                                    
\(=\frac{10^9-2}{10^9-2}+\frac{7}{10^9-2}\)

\(=1+\frac{7}{10^9-2}\)

\(B=\frac{10^9}{10^9-7}\)

\(=\frac{10^9-7}{10^9-7}+\frac{7}{10^9-7}\)

\(=1+\frac{7}{10^9-7}\)

Vì \(7\over10^9-5\)<\(7\over10^9-7\) nên A<B

a )  Ta có : 

\(\frac{9^{10}-4}{9^{10}-5}=\frac{9^{10}-5+1}{9^{10}-5}=1+\frac{1}{9^{10}-5}\)

\(\frac{9^{10}-2}{9^{10}-3}=\frac{9^{10}-3+1}{9^{10}-3}=1+\frac{1}{9^{10}-3}\)

Do \(\frac{1}{9^{10}-5}>\frac{1}{9^{10}-3}\)

\(\Rightarrow1+\frac{1}{9^{10}-5}>1+\frac{1}{9^{10}-3}\)

\(\Rightarrow\frac{9^{10}-4}{9^{10}-5}>\frac{9^{10}-2}{9^{10}-3}\)

b ) Ta có : 

\(\frac{2.7^{10}-1}{7^{10}}=2-\frac{1}{7^{10}}\)

\(\frac{2.7^{10}+1}{7^{10}+1}=\frac{2.7^{10}+2-1}{7^{10}+1}=\frac{2\left(7^{10}+1\right)-1}{7^{10}+1}=2-\frac{1}{7^{10}+1}\)

Do \(\frac{1}{7^{10}}>\frac{1}{7^{10}+1}\)

\(\Rightarrow2-\frac{1}{7^{10}}< 2-\frac{1}{7^{10}+1}\)

\(\Rightarrow\frac{2.7^{10}-1}{7^{10}}< \frac{2.7^{10}+1}{7^{10}+1}\)

mình xem chả hiểu đây này

+ta có 10^2010=10...0(2010 số 0)

và 10^2011=10...0(2011 số 0)

suy ra  -9/10...0(2010 số 0)= -90/10...0(2011 số 0)[nhân tử,mẫu cho 10]

suy ra A=-90/10...0(2011 số 0)+-19/10...0(2011 số 0)= -109/10...0(2011 số 0)     [1]

+-19/10...0(2010 số 0)= -190/10...0(2011 số 0)[nhân tử,mẫu cho 10]

và 10^2011=10...0(2011 số 0)

suy ra -9/10...0(2011 số 0)+-190/10...0(2011 số 0)= -199/10...0(2011 số 0)    [2]

vì -109>-199 suy ra [1]>[2]

K CHO MIK VS BẠN ƠIIIIIIIIIIIIIIIIIII

\(-A=\frac{9}{10^{2010}}+\frac{19}{10^{2011}}\)

\(-A=\frac{9}{10^{2010}}+\frac{10}{10^{2011}}+\frac{9}{10^{2011}}\)

\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)

\(-A=\frac{10}{10^{2010}}+\frac{9}{10^{2011}}\)

\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)

\(-B=\frac{9}{10^{2011}}+\frac{19}{10^{2010}}\)

Làm tương tự nhé 

ta thấy -b > -a nên a>b

yêu cầu so sánh 2 phân số

Đặt \(A=\frac{10^{2006}+9}{10^{2007}+9}\)

\(\Rightarrow10A=\frac{10^{2007}+90}{10^{2007}+9}=1+\frac{81}{10^{2007}+9}\)

\(\frac{10^{2007}+9}{10^{2008}+9}=B\)

\(\Rightarrow10B=\frac{10^{2008}+90}{10^{2008}+9}=1+\frac{81}{10^{2008}+9}\)

Vì\(10A>10B\Rightarrow A>B\)

Đặt \(A=\frac{10^{2006}+9}{10^{2007}+9}\) và \(B=\frac{10^{2007}+9}{10^{2008}+9}\)

* Cách 1 : 

Ta có : 

\(10A=\frac{10^{2007}+90}{10^{2007}+9}=\frac{10^{2007}+9+81}{10^{2007}+9}=\frac{10^{2007}+9}{10^{2007}+9}+\frac{81}{10^{2007}+9}=1+\frac{81}{10^{2007}+9}\)

\(10B=\frac{10^{2008}+90}{10^{2008}+9}=\frac{10^{2008}+9+81}{10^{2008}+9}=\frac{10^{2008}+9}{10^{2008}+9}+\frac{81}{10^{2008}+9}=1+\frac{81}{10^{2008}+9}\)

Vì \(\frac{81}{10^{2007}+9}>\frac{81}{10^{2008}+9}\) nên \(1+\frac{81}{10^{2007}+9}>1+\frac{81}{10^{2008}+9}\)

Vậy \(A>B\)

* Cách 2 : 

Ta có công thức : 

\(\frac{a}{b}< \frac{a+m}{b+m}\) \(\left(\frac{a}{b}< 1;a,b,m\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2007}+9}{10^{2008}+9}< \frac{10^{2007}+9+1}{10^{2008}+9+1}=\frac{10^{2007}+10}{10^{2008}+10}=\frac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\frac{10^{2006}+1}{10^{2007}+1}=A\)

Ta thấy \(B< A\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

a) Ta có:

\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)

\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)

Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)

b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)

Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)

- Ta có:

\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)

- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)

a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

=>Bằng nhau

b: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)

ta có : A = \(\frac{7^{10}}{1+7+7^2+7^3+...+7^9}=1:\frac{1+7+7^2+7^3+...+7^9}{7^{10}}\)

= \(1:\left(\frac{1}{7^{10}}+\frac{7}{7^{10}}+\frac{7^2}{7^{10}}+...+\frac{7^8}{7^{10}}+\frac{7^9}{7^{10}}\right)\)=\(1:\left(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\right)\)

tương tự ta được : B = \(1:\left(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\right)\)

Vì \(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\)< \(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\)

=> A > B 

a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)

Đặt \(B=1+7+7^2+...+7^{14}\)

\(\Rightarrow7B=7+7^2+...+7^{15}\)

\(\Rightarrow7B-B=6B=7^{15}-1\)

\(\Rightarrow B=\frac{7^{15}-1}{6}\)

\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)

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