Ta có:  \(\left(x-\frac{1}{5}\right)^{2020}\ge0\forall x\)

         \(\left(y+0.4\right)^{2000}\ge0\forall y\)

        \(\left(z-3\right)^6\ge0\forall z\)

=> \(\left(x-\frac{1}{5}\right)^{2020}+\left(y+0.4\right)^{2000}+\left(z-3\right)^6\ge0\forall x,y,z\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-\frac{1}{5}=0\\y+0.4=0\\z-3=0\end{cases}}\) => \(\hept{\begin{cases}x=\frac{1}{5}\\y=0\\z=3\end{cases}}\)

vậy ...

Ư(8)={1;2;4;8}

Vậy có 4 số là ước của 8

 8 = 2³ 

Số U của 8 là : 3 + 1 = 4 U 

Vậy cs 4 số tự nhiên là U của 8

 Hc tốt

\(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\)\(\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

\(=\left(1+\frac{\sqrt{a}}{a+1}\right):\)\(\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)-\left(a+1\right)}\right)\)

\(=\left(1+\frac{\sqrt{a}}{a+1}\right):\)\(\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\left(1+\frac{\sqrt{a}}{a+1}\right):\)\(\left(\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\left(\frac{a+\sqrt{a}+1}{a+1}\right):\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{\left(a+\sqrt{a}+1\right)\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{\left(a+\sqrt{a}+1\right)}{a+1}\)

x = 67569

VÌ 20192019+120192020 +1=140384040 >20192018+120192019 =140384038 nên A>B

N > M nhé bạn.

a) Đặt :\(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\)

\(\Rightarrow a=kb;c=kd\)

Thay: \(\frac{a^2-b^2}{c^2-d^2}=\frac{k^2.b^2-b^2}{k^2.d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\) ⁽¹⁾

Mặt khác: \(\frac{ab}{cd}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\) 

b) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\) ⁽³⁾

Từ (2) và (3) \(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\left(đpcm\right)\)

a.

Đặt \(k=\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{k^2b^2-b^2}{k^2d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}=\frac{b}{d}\cdot\frac{b}{d}\)

Mà \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Thay vào ta được \(đpcm\)

b tương tự 

1

\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)

\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

\(=\frac{100^{100}+1}{100^{99}+1}=N\)