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a) Ta có: a < b => a + 1 < b + 1
b) Ta có: a < b => a - 2 < b - 2
1.
Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\Leftrightarrow ab+ad< ad+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
Lại có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
2.
Ta có: a(b + n) = ab + an (1)
b(a + n) = ab + bn (2)
Trường hợp 1: nếu a < b mà n > 0 thì an < bn (3)
Từ (1),(2),(3) suy ra a(b + n) < b(a + n) => \(\frac{a}{n}< \frac{a+n}{b+n}\)
Trường hợp 2: nếu a > b mà n > 0 thì an > bn (4)
Từ (1),(2),(4) suy ra a(b + n) > b(a + n) => \(\frac{a}{b}>\frac{a+n}{b+n}\)
Trường hợp 3: nếu a = b thì \(\frac{a}{b}=\frac{a+n}{b+n}=1\)
\(C=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow D< 1-\frac{1}{2017}< 1\)
Vậy C > D
1)
\(\frac{a}{b}=\frac{a\left(b+c\right)}{b\left(b+c\right)}=\frac{ab+ac}{b\left(b+c\right)}\)
\(\frac{a+c}{b+c}=\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+bc}{b\left(b+c\right)}\)
mà ab = ab; ac > bc ( vì a > b )
=> \(\frac{a}{b}>\frac{a+c}{b+c}\left(đpcm\right)\)
DO a,b,c đối xứng , giả sử \(a\ge b\ge c\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\end{cases}}\)
áp dụng bất đẳng thức trê-bư-sép ta có
\(a^2.\frac{a}{b+c}+b^2.\frac{b}{a+c}+c^2.\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{3}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=\frac{1}{3}.\frac{3}{2}=\frac{1}{2}\)
vậy \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{2}\)dấu bằng xảy ra khi\(a=b=c=\frac{1}{\sqrt{3}}\)
\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)
\(\Rightarrow S=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{abc}{abc+c.abc+ca}\)
\(S=\frac{abc}{a.\left(bc+b+1\right)}+\frac{1}{1+b+bc}+\frac{abc}{ac.\left(bc+b+1\right)}\)
\(S=\frac{bc}{bc+b+1}+\frac{1}{1+b+bc}+\frac{b}{bc+b+1}\)
\(S=\frac{bc+b+1}{bc+b+1}\)
\(S=1\)
Điều kiện \(c\ge0\);\(a;b>0\)
Ta có: \(a>b\)
\(\Rightarrow ac\ge bc\)
\(\Rightarrow ac+ab\ge bc+ab\)
\(a.\left(b+c\right)\ge b.\left(c+a\right)\)
\(\Rightarrow\frac{a+c}{b+c}\ge\frac{a}{b}\)
Tham khảo nhé~
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
\(\frac{a^2}{1+a+a^2}\)
\(\frac{1}{1+a}\)
\(\frac{b^2}{1+b+b^2}\)=\(\frac{1}{1+b}\)
vì a>b nên \(\frac{a^2}{1+a+a^2}\)>\(\frac{b^2}{1+b+b^2}\)