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a) Xét:
\(a>b\)
\(\Rightarrow\dfrac{a}{b}>1\Rightarrow\dfrac{a+m}{b+m}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{a+m}\)
\(a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
\(a=b\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow\dfrac{a+m}{b+m}=1\Rightarrow\dfrac{a}{b}=\dfrac{a+m}{b+m}=1\)
Mk chỉ áp dụng tính 1 câu,câu sau làm tương tự
b)
Ta có:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{1993}+1}{10^{1992}+1}< 1\)
\(B< \dfrac{10^{1993}+1+9}{10^{1992}+1+9}\Rightarrow B< \dfrac{10^{1993}+10}{10^{1992}+10}\Rightarrow B< \dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\Rightarrow B< \dfrac{10^{1992}+1}{10^{1991}+1}=A\)
\(B< A\)
@@ ~ học tốt ~
\(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-90}{10^{2011}}+\dfrac{-19}{10^{2011}}=\dfrac{\left(-90\right)+\left(-19\right)}{10^{2011}}=\dfrac{-109}{10^{2011}}\)\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-190}{10^{2011}}=\dfrac{\left(-9\right)+\left(-190\right)}{10^{2011}}=\dfrac{-199}{10^{2011}}\)\(\text{Vì }\dfrac{-109}{10^{2011}}>\dfrac{-199}{10^{2011}}\text{ nên }A>B\)
1.
ta có: 2009A= (2009^2010+ 2009)/ (2009^2010+1)= (2009^10+1+2008)/(2009^2010+1)=1+ [2008/(2009^2010+1)]
làm tương tự như trên ta được :
2009B=1-[4016/(2009^2011-2)]
lại có:
2009A= .............(nt) > 1
2009B=...........<1
=>2009A>2009B
=>A>B
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
Ta có: \(10A=\dfrac{10^{2016}-10}{10^{2016}-1}=1-\dfrac{9}{10^{2016}-1}\)
\(10B=\dfrac{10^{2015}+10}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)
Vì \(\dfrac{9}{10^{2016}-1}< \dfrac{9}{10^{2015}+1}\Rightarrow1-\dfrac{9}{10^{2016}-1}< 1+\dfrac{9}{10^{2015}+1}\)
\(\Rightarrow10A< 10B\Rightarrow A< B\)
Vậy A < B
Ta có: \(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2011}}\)
\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-9}{10^{2010}}+\dfrac{-10}{10^{2010}}\)
So sánh A với B ta thấy: \(\dfrac{-9}{10^{2010}}=\dfrac{-9}{10^{2010}};\dfrac{-9}{10^{2011}}=\dfrac{-9}{10^{2011}}\)
Mà \(\dfrac{-10}{10^{2011}}>\dfrac{-10}{10^{2010}}\)
\(\Rightarrow\) \(\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2011}}>\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2010}}\)
\(\Rightarrow\) \(A>B\)
Vậy A > B.
A=\(\dfrac{2009^{2010}+1}{2009^{2009}+1}\)
2009A=\(\dfrac{(2009^{2010}+1)+0}{2009^{2010}+1}\)
= 1+\(\dfrac{0}{2009^{2010}+1}\)= 1+0 =1
B=\(\dfrac{2009^{2011}-2}{2009^{2010}-2}\)
2009B=\(\dfrac{2009^{2011}-1}{2009^{2011}-2009}\)
=\(\dfrac{(2009^{2011}-1)-0}{2009^{2011}-2009}\)
= \(1-\dfrac{0}{2009^{2011}-2009}\)
=1-0= 1
Vì 1=1\(\Rightarrow A=B\)
Ta có : A = 2009^2010+1/2009^2009+1
Suy ra: 1/2009 A = 1 - 2008/2009^2010+2009 (1)
Lại có:B = 2009^2011 - 2 / 2009^2010 - 2
Suy ra : 1/2009 B = 1 + 4016/2009^2011-4018 (2)
Vì 1 - 2008/2009^2010+2009 < 1 + 4016/2009^2011-4018 (3)
Từ (1);(2) và (3) suy ra : A<B
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
A=-2015/2015x2016
A=-1/2016
B=-2014/2014x2015
B=-1/2015
vi 2016>2015,-1/2016>-1/2015
vay A>B
b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)
Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)
Ta có: \(10^{2010}+1< 10^{2011}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)
\(\Leftrightarrow10A>10B\)
hay A>B