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\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\)

\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\)(*)

Đặt \(x^2-5x+4=a\)

(*)\(\Leftrightarrow a\left(a+2\right)\)

\(=a^2+2a\)

\(=a^2+2a+1-1\)

\(=\left(a+1\right)^2-1\ge-1\forall a\)

Dấu "=" xảy ra \(\Leftrightarrow a=-1\Leftrightarrow x^2-5x+4=-1\)

\(\Leftrightarrow x^2-5x+5=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\left(\frac{\pm\sqrt{5}}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+5}{2}\\x=\frac{-\sqrt{5}+5}{2}\end{cases}}\)

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)

\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)

\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

Vì 20 < 21 nên 11/20 > 11/21

Vậy ..... 

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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\)

\(A=\frac{1}{20}\)

Vì 20 < 21 nên 1/20 > 1/21 

Vậy ............

\(\frac{1}{x+1}\)và \(\frac{x+2}{2x+1}\)

Ta có :

\(\frac{1}{x+1}=\frac{1.2x+1}{x+1.2x+1}=\frac{2x+1}{3x+1}\)

\(\frac{x+2}{2x+1}=\frac{x+2.x+1}{2x+1.x+1}=\frac{3x+1}{3x+1}\)

Vì \(\frac{2x+1}{3x+1}< \frac{3x+1}{3x+1}\)

=> \(\frac{1}{x+1}< \frac{x+2}{2x+1}\)

Vậy :...

\(P=\dfrac{1+\sqrt{x}}{2\sqrt{x}}=\dfrac{\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{x}}\)

Do \(\dfrac{1}{2\sqrt{x}}>0\)

\(\Rightarrow P=\dfrac{1}{2}+\dfrac{1}{2\sqrt{x}}>\dfrac{1}{2}\)

(x-1)²⁰²⁰=(x-1)²⁰²²

=>(x-1)²⁰²⁰-(x-1)²⁰²²=0

=>(x-1)²⁰²⁰-(x-1)²⁰²⁰.(x-1)²=0

=>(x-1)²⁰²⁰(1-(x-1)²=0

=>(x-1)²⁰²⁰=0 hoặc 1-(x-1)²=0

=>x=1 hoặc x=2.

Bài 2

a,2¹⁰⁵ và 5⁴⁵

2¹⁰⁵=(2⁷)¹⁵=128¹⁵

5⁴⁵=(5³)¹⁵=125¹⁵

Vì 128^15_>12515 nên 2¹⁰⁵>5⁴⁵.

b,

5⁵⁴ và 3⁸¹

5⁵⁴=(5⁶)⁹=15625⁹

3⁸¹=(3⁹)⁹=19683⁹

Vì 15625⁹<19683⁹ nên 5⁵⁴<3⁸¹

Bài 1 :

\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)

\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)

\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)

\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)

\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)

\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)

\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)

\(A=\frac{-2015}{4028}\)

a)  Có \(x+1< x+2\)

\(\Rightarrow\sqrt{x+1}< \sqrt{x+2}\)

\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+2}}< 1\)

b)  Vì \(\sqrt{x+1}< \sqrt{x+2}\)

\(\Rightarrow\sqrt{x+1}.\sqrt{x+1}.\sqrt{x+2}< \sqrt{x+2}.\sqrt{x+1}.\sqrt{x+1}\)

\(\Leftrightarrow\sqrt{x+1}^2.\sqrt{x+2}< \sqrt{x+2}^2.\sqrt{x+1}\)

\(\Rightarrow\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}< \frac{\sqrt{x+1}}{\sqrt{x+2}}\)

hay \(\frac{\sqrt{x+1}}{\sqrt{x+2}}>\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}\)

a: Ta có: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b:Để M=2 thì \(\sqrt{x}-1=2\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=-1\left(loại\right)\)