546/4 nha e 

hoặc là 273/2 c ngĩ là 273/2 nó tối giẩn nên ghi đáp án là 273/2 nhé

15 nha

3 phần4 nhân 20

a, Theo định lí Pytago tam giác ADE vuông tại A

\(DE=\sqrt{AD^2+AE^2}=20cm\)

b, Ta có DE > AD > AE 

=> ^A > ^E > ^D 

c, Ta có \(S_{AED}=\dfrac{1}{2}.AD.AE;S_{AED}=\dfrac{1}{2}.AH.DE\)

\(\Rightarrow AH=\dfrac{AD.AE}{DE}=\dfrac{48}{5}cm\)

Theo định lí Pytago tam giác ADH vuông tại H

\(DH=\sqrt{AD^2-AH^2}=\dfrac{64}{5}cm\)

=> DH > AH 

Trả lời:

a) \(\frac{10^{2005}+1}{10^{2006}+1}\) và \(\frac{10^{2004}+1}{10^{2005}+1}\)

Đặt \(A=\frac{10^{2005}+1}{10^{2006}+1}\)\(;\)\(B=\frac{10^{2004}+1}{10^{2005}+1}\)

\(+)\)\(A=\frac{10^{2005}+1}{10^{2006}+1}\)

\(\Rightarrow10A=\frac{10^{2006}+10}{10^{2006}+1}\)

\(\Rightarrow10A=\frac{(10^{2006}+1)+9}{10^{2006}+1}\)

\(\Rightarrow10A=1+\frac{9}{10^{2006}+1}\)

\(+)\)\(B=\frac{10^{2004}+1}{10^{2005}+1}\)

\(\Rightarrow10B=\frac{10^{2005}+10}{10^{2005}+1}\)

\(\Rightarrow10B=\frac{(10^{2005}+1)+9}{10^{2005}+1}\)

\(\Rightarrow10B=\frac{9}{10^{2005}+1}\)

Vì \(\frac{9}{10^{2006}+1}\)\(< \)\(\frac{9}{10^{2005}+1}\)\(;\) \(1=1\)

\(\Rightarrow\)\(1+\frac{9}{10^{2006}+1}\) \(< \)\(1+\frac{9}{10^{2005}+1}\)

\(\Rightarrow\)\(A< B\)

Hay \(\frac{10^{2005}+1}{10^{2006}+1}\)\(< \)\(\frac{10^{2004}+1}{10^{2005}+1}\)

Vậy...

43 đoán thế hihi

cảm ơn nhé

4/5 : 12/60 = 4 nha

HT

20 đó tị nhs

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1.884m nha