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A = 1 + 2 + 3 +......+ x = 55
<=> (1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5 +....+ x = 55
<=> x = 55 - [(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5 +....]
<=> x = 55 - (45 + ...)
<=> x = 10 - (....)
=> x = 10
A = 1 + 2 + 3 +......+ x = 55
<=> (1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5 +....+ x = 55
<=> x = 55 - [(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5 +....]
<=> x = 55 - (45 + ...)
<=> x = 10 - (....)
Nếu x > 0 => x = 10
___ x < 0 => x = {-10; -20;....}
( 3/7 + 1/2 ) 2 = 3/72+2x3/7x1/2+1/22=169/196
k minh ban nha 243254354543333333351343545345325454353425435423454
( 3/7 + 1/2 ) ^2 = ( 13/14)^2= 169/196
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)(\(a;b;m\in\)N*)
Ta có:
\(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)
\(B< \frac{10^{2007}+10}{10^{2008}+10}\)
\(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)
\(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)
=> \(B< A\)
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+..+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\left(đpcm\right)\)
\(\frac{b}{a+b}=\frac{c}{b+c}=\frac{a}{a+c}\Rightarrow\frac{a+b}{b}=\frac{b+c}{c}=\frac{a+c}{a}\)
\(\Leftrightarrow\frac{a}{b}+1=\frac{b}{c}+1=\frac{c}{a}+1\)mà\(a,b,c>0\Rightarrow a+b+c\ne0\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
