1: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b+c}{2+3+4}=\dfrac{180}{9}=20\)

Do đó: a=40; b=60; c=80

Xét ΔABC có \(\widehat{A}< \widehat{B}< \widehat{C}\)

nen BC<AC<AB

2: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{4}}=\dfrac{b+c}{\dfrac{1}{3}+\dfrac{1}{4}}=\dfrac{70}{\dfrac{7}{12}}=120\)

Do đó: b=40; c=30

Xét ΔABC có \(\widehat{A}>\widehat{B}>\widehat{C}\)

nên BC>AC>AB

Ý của bài là:

A = 2⁹¹

3³⁵ 

Hãy so sánh A và 3³⁵

2⁹¹ = ( 2¹³ )⁷ = 8082⁷

3³⁵ = ( 3⁵ )⁷ = 243⁷

Mà 8082⁷ > 243⁷ =>A > 3³⁵ 

A ? A là gì ?

Ta có:

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...

\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)

\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)

\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)

\(\Rightarrow P< 1\)

Ta có :

\(\frac{2015.2000-15}{2016.1999+1}\)

= \(\frac{2015.1999+2015-15}{2015.1999+1999+1}\)

= \(\frac{2015.1999+2000}{2015.1999+2000}\)

= 1

Vậy \(\frac{2015.2000-15}{2016.1999+1}=1\)

\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}

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\(a^2\ge a\)

Nếu \(a>1\Rightarrow a^2>a\)

Nếu\(1>a>0\Rightarrow a^2< a\)

Nếu\(a< 0\Rightarrow a^2>a\)

Nếu\(a=0;1\Rightarrow a^2=a\)