a ) Ta có : \(9^{20}\)= \(\left(3^2\right)^{10}\)= \(3^{20}\)

                \(27^{13}\)= \(\left(3^3\right)^{13}\)= \(3^{39}\)

Vì 39 > 20 => 9^ 20 < 27 ^ 13

Phần b bạn vào câu hỏi tương tự. Nhớ tích đúng cho tớ

A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1 => \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\) = \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\) = \(\frac{2005^{2004}+1}{2005^{2005}+1}\) = B => A<B.

Ta thấy:A=\(\frac{2005^{2005+1}}{2005^{2006}+1}\)<1
Ta có:A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)<\(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)=\(\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=b
Vậy A<B
Chắc chắn 100%

cái này trong đề cương

a.\(\frac{13}{15}< \frac{23}{25}\)

a.\(\frac{25}{18}>\frac{24}{27}\)

c.\(\frac{25}{78}< \frac{24}{27}\)

d.\(\frac{13}{15}< \frac{133}{153}\)

e.\(\frac{2003}{2004}+\frac{2004}{2005}< 2003+\frac{2004}{2004}+2003\)

1.\(a,\)Ta có: \(1-\frac{13}{15}=\frac{2}{15};1-\frac{23}{25}=\frac{2}{25}\)

Mà \(\frac{2}{15}>\frac{2}{25}\)

Vì nếu cùng số bị trừ, số trừ càng lớn thì thương cang nhỏ và ngược lại. Do cùng bị 1 trừ nên \(\frac{13}{15}< \frac{23}{25}\)

\(b,\)Ta có: \(\frac{24}{27}=\frac{8}{9}=\frac{16}{18}\)

Mà \(\frac{25}{18}>\frac{16}{18}\)

Nên \(\frac{25}{18}>\frac{24}{27}\)

\(c,\)Ta có: \(\frac{24}{27}=\frac{8}{9}\)

Và \(\frac{25}{78}=\frac{25.9}{78.9};\frac{8}{9}=\frac{8.78}{9.78}\)

Mà \(25.9=25\left(8+1\right)=25.8+25< 8.78\)

Nên \(\frac{25}{78}< \frac{8}{9}=\frac{24}{27}\)

\(d,\)Ta có: \(1-\frac{13}{15}=\frac{2}{15}=\frac{20}{150}\)

\(1-\frac{133}{153}=\frac{20}{153}>\frac{20}{150}=\frac{2}{15}\)

Vì nếu cùng số bị trừ, số trừ càng lớn thì thương cang nhỏ và ngược lại. Do cùng bị 1 trừ nên \(\frac{133}{153}< \frac{13}{15}\)

2.        Ta có: \(\frac{2003+2004}{2004+2003}=\frac{2007}{2007}=1\)

Còn tiếp nữa thì bạn tự giải nha! chỉ cần so sánh 2003/2004+2004/2005 với 1 thôi!

bài hai bạn có thể làm cách hai như sau:

Ta có: \(\frac{2003}{2004}>\frac{2003}{2007}=\frac{2003}{2004+2003};\frac{2004}{2005}>\frac{2004}{2007}=\frac{2004}{2004+2003}\)

Suy ra: \(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003}{2004+2003}+\frac{2004}{2004+2003}=\frac{2003+2004}{2004+2003}\)

Vậy: \(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2003}\)

nhớ k cho mình nhé

a)A = B

b)A>B

$A=\frac{{2005}^{2005}+1}{{2005}^{2006}+1}<\frac{{2005}^{2005}+1+2004}{{2005}^{2006}+1+2004}=\frac{2005.({2005}^{2004}+1)}{2005.({2005}^{2005}+1)}==\frac{{2005}^{2004}+1}{{2005}^{2005}+1}=B.$

Vậy A < B

$A=\frac{{2005}^{2005}+1}{{2005}^{2006}+1}<\frac{{2005}^{2005}+1+2004}{{2005}^{2006}+1+2004}=\frac{2005.({2005}^{2004}+1)}{2005.({2005}^{2005}+1)}==\frac{{2005}^{2004}+1}{{2005}^{2005}+1}=B.$

Vậy A < B

$A=\frac{{2005}^{2005}+1}{{2005}^{2006}+1}<\frac{{2005}^{2005}+1+2004}{{2005}^{2006}+1+2004}=\frac{2005.({2005}^{2004}+1)}{2005.({2005}^{2005}+1)}==\frac{{2005}^{2004}+1}{{2005}^{2005}+1}=B.$

Vậy A < B

CHÚC HỌC TỐT VÀ ĐỪNG QUÊN TICK NHA

\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

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

vì 2005²⁰⁰⁶+1>2005²⁰⁰⁵+1

\(\Rightarrow\frac{4}{2005^{2006}+1}< \frac{4}{2005^{2005}+1}\)

\(\Rightarrow1+\frac{4}{2005^{2006}+1}< 1+\frac{4}{2005^{2005}+1}\)

=>A<B

sai đề bài

uk sai ở mẫu của A

2) \(-x^2+4x-2\)

\(=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)\)

\(=-\left(x-2\right)^2+2\)

Ta có: \(-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)

Vậy: GTLN của bt là 2 tại x=2

b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))

Mà: \(\sqrt{2x^2-3}\ge0\forall x\)

Dấu "=" xảy ra:

\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)

Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)

...

1:

b: \(4\sqrt{5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{75}\)

=>\(4\sqrt{5}>5\sqrt{3}\)

=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)

c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)

=>\(3-2\sqrt{5}< 1-\sqrt{5}\)

d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)

=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)

e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)

\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)

=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)

=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)

Ta có VẾ A

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

\(2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}\)

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

\(2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)

\(2005\cdot A=1+\frac{2004}{2005^{2006}+1}\)

Ta lại có Vế B :

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

\(2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}\)

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

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

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

Nhìn vào trên , suy ra A < B . 

\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

\(\Rightarrow A< B\)