\(A=2^1-2^2+...+2^{2011}-2^{2012}\)

\(\Rightarrow2A=2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow A+2A=\left(2^1-2^2+...+2^{2011}-2^{2012}\right)+\left(2^2-2^3+...+2^{2012}-2^{2013}\right)\)

\(\Rightarrow3A=2-2^{2013}\)

\(\Rightarrow A=\dfrac{2-2^{2012}}{3}\)

Ta có:

\(\dfrac{1}{51}>\dfrac{1}{100},\dfrac{1}{52}>\dfrac{1}{100},...,\dfrac{1}{100}=\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\left(1\right)\)

\(\dfrac{1}{51}< \dfrac{1}{50},\dfrac{1}{52}< \dfrac{1}{50},...,\dfrac{1}{100}< \dfrac{1}{50}\)

\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}.50=1\left(2\right)\)

Từ (1),(2)\(\RightarrowĐPCM\)

Gọi độ dài 3 cạnh là a,b,c; 3 chiều cao tương ứng là x,y,z .Diện tích là S

Ta có :\(a=\dfrac{2S}{x},b=\dfrac{2S}{y},c=\dfrac{2S}{z}\)

Mà \(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{c}{6}\)

\(\Rightarrow\dfrac{2S}{4x}=\dfrac{2S}{5y}=\dfrac{2S}{6z}\)

\(\Rightarrow4x=5y=6z\)

\(\Rightarrow\dfrac{4x}{60}=\dfrac{5y}{60}=\dfrac{6z}{60}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{12}=\dfrac{z}{10}\)

Vậy 3 chiều cao tương ứng tỉ lệ với 15,12,10

1. Trang had (have) a big breakfast this morning.She ate (eat)a loaf of bread and three eggs.

2.Ats the moment Thoa is sitting (sit) in Dr.Khoa's surgery.He will check (check)her teeth.

3.How often does your uncle go (go) to she hospital to have (have) a medical check-up?

-He goes(go) there twice a year.

4.Take these pills and you will feel (feel) better then,Ba!

5.Minh didn't visit (not visit) the museum with his class last Sunday because he caught (catch) a cold.

Ta thấy:

\(\sqrt{50}>\sqrt{49}=7\)

\(\sqrt{26}>\sqrt{25}=5\)

\(\Rightarrow\sqrt{50}+\sqrt{26}+1>7+5+1=13\)(1)

Mà \(\sqrt{168}< \sqrt{169}=13\left(2\right)\)

Từ (1),(2)\(\Rightarrow\sqrt{50}+\sqrt{26}+1>\sqrt{168}\)

1)\(\dfrac{x+1}{-12}=\dfrac{-3}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy....

b)\(\left(\dfrac{1}{2}-2^2:\dfrac{4}{3}\right).\dfrac{6}{5}-7\)

\(=\left(\dfrac{1}{2}-4.\dfrac{3}{4}\right).\dfrac{6}{5}-7\)

\(=\left(\dfrac{1}{2}-3\right).\dfrac{6}{5}-7\)

\(=\dfrac{-5}{2}.\dfrac{6}{5}-7\)

\(=-3-7\)

\(=-10\)

sau tất cả ...(mk chỉ nhớ có ậy thui)

ngày mai mk làm cho

Ta có :\(\dfrac{a}{a+b+c}< \dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\)

\(\dfrac{b}{a+b+c}< \dfrac{b}{c+b}< \dfrac{b+a}{a+b+c}\)

\(\dfrac{c}{a+b+c}< \dfrac{c}{a+c}< \dfrac{b+c}{a+b+c}\)

\(\Rightarrow\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}\)\(\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

\(\Rightarrow\)ĐPCM