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a. Ta có
\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}.\)
Vì\(\frac{2011}{2012+2013}< \frac{2011}{2012}.\)(1)
\(\frac{2012}{2012+2013}< \frac{2012}{2013}.\)(2)
Cộng vế với vế của 1;2 ta được
\(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}< A=\frac{2011}{2012}+\frac{2012}{2013}\)
hay A>B
~.~
M lớn hơn hay nhỏ hơn N vậy bạn ơi??
Nếu m > n thì A > B; m < n thì A < B nhé!!
TA CÓ :
\(B=\frac{2010+2011+2012}{2011+2012+2013}\)
\(B=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
VÌ : \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
=> A > B
VẬY , A > B
Mình tự hỏi. sao banh biết rồi còn đăng lên làm gì??????????
Ta có
\(\frac{A^{2011}}{A^{2012}}=\frac{A^{2012}}{A^{2103}}=\frac{A}{A^2}\)
=> \(\frac{A^{2011}}{A^{2012}}+\frac{A^{2012}}{A^{2013}}=\frac{2A}{A^2}\)
\(\frac{A^{2011+2012}}{A^{2012+2013}}=\frac{A^{4023}}{A^{4025}}=\frac{1}{A^2}\)
=> \(\frac{A^{2011+2012}}{A^{2012+2013}}< \frac{A^{2011}}{A^{2012}}+\frac{A^{2012}}{A^{2013}}\)
Áp dụng BĐT \(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}>\frac{a+b+c}{a+b+c}=1>\frac{a+b+c}{b+c+d}\).
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2010+2011+2012}>\frac{2010+2011+2012}{2011+2012+2013}\)mà 2010 + 2011 + 2012 < 2011+2012+2013 ,suy ra \(\frac{2010+2011+2012}{2011+2012+2013}< 1\))
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2011+2012+2013}\)hay P > Q
Vậy P > Q
b) Áp dụng công thức BCNN (a, b) . UCLN (a,b) = a.b
\(\Rightarrow a.b=420.21=8820\)
Ta có:
\(ab=8820\)
\(a+21=b\Rightarrow b-a=21\)
Hai số cách nhau 21 mà có tích là 8820 là 84 , 105
Mà a + 21 = b suy ra a < b
Vậy a = 84 ; b = 105
a,-Cách khác:
-Ta có: \(\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
-Mà: \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\left(1\right)\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\left(2\right)\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\left(3\right)\)
\(\Rightarrow P>Q\)
a ) Nếu \(\frac{a}{b}>\frac{a+m}{b+m}\)
\(\Leftrightarrow a\left(b+m\right)>b\left(a+m\right)\)
\(\Leftrightarrow ab+am>ab+bm\)
\(\Leftrightarrow am>bm\)
\(\Rightarrow a>b\)
\(\Rightarrow\frac{a}{b}>1\)
Vậy \(\frac{a}{b}>1\) thì \(\frac{a}{b}>\frac{a+m}{b+m}\)
b ) Vì 237 > 142 => \(\frac{237}{142}>\frac{237+9}{142+9}=\frac{246}{151}\)
Xét hiệu :
\(\frac{a}{b}-\frac{a+m}{b+m}\)
\(=\frac{a\left(b+m\right)}{b\left(b+m\right)}-\frac{\left(a+m\right)b}{\left(b+m\right)b}\)
\(=\frac{a.b+a.m}{b\left(b+m\right)}-\frac{a.b+b.m}{b\left(b+m\right)}\)
\(=\frac{a.b+a.m-a.b+b.m}{b\left(b+m\right)}\)
\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}\)
Vì \(\frac{a}{b}>1,b\in\)N* \(\Rightarrow a>b\Rightarrow a-b>0,m\in\)N*
\(\Rightarrow m\left(a-b\right)>0\); Vì : \(b,m\in\)N* \(\Rightarrow b\left(b+m\right)>0\)
\(\Rightarrow\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\) hay : \(\frac{a}{b}-\frac{a+m}{b+m}>0\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
Vậy \(\frac{a}{b}>1,m\in\)N* thì \(\frac{a}{b}>\frac{a+m}{b+m}\)
b, Tự làm
Ta có \(B=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}\)
Lại có: \(\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}\) ( ngoặc 2 dòng này lại nhé dòng này và dòng trên)
\(\Rightarrow B>A\)
ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013};\frac{2012}{2013}>\frac{2012}{2013+2012}.\)
\(\Rightarrow A>\frac{2011}{2012+2013}+\frac{2012}{2013+2012}=\frac{2011+2012}{2012+2013}=B\)
....
Ta có \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
CỘNG VẾ THEO VẾ,TA CÓ:
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011+2012}{2012+2013}\)
\(\Rightarrow A>B\)
Vậy A>B
Gọi 2011 là a
2012 là b;2013 là c
=>\(A=\frac{2011}{2012}+\frac{2012}{2013}=\frac{a}{b}+\frac{b}{c}\);\(B=\frac{2011+2013}{2012+2013}=\frac{a+c}{b+c}\)
=>\(A=\frac{a}{b}+\frac{b}{c}=\frac{ac+b^2}{bc}\)\(=\frac{\left(ac+b^2\right).\left(b+c\right)}{bc.\left(b+c\right)}\);\(B=\frac{a+c}{b+c}=\frac{\left(a+c\right).bc}{bc.\left(b+c\right)}\)
b+c>a+c;b2+ac>bc
Vậy A>B
Xét hiệu \(A-B=\frac{2013-2012}{a^n}+\frac{2011-2012}{a^m}=\frac{1}{a^n}-\frac{1}{a^m}\)
TH1: n > m > 0
=> an > am \(\Rightarrow\frac{1}{a^n}<\frac{1}{a^m}\Rightarrow\frac{1}{a^n}-\frac{1}{a^m}<0\)=> A < B
TH2: m > n > 0
=> an < am \(\Rightarrow\frac{1}{a^n}>\frac{1}{a^m}\Rightarrow\frac{1}{a^n}-\frac{1}{a^m}>0\Rightarrow A>B\)
Đề sai, sao lại có x ở đây?