Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)

3:

a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

2:

a: 2căn 7=căn 28

3căn 2=căn 18

mà 28>18

nên 2*căn 7>3*căn 2

b: 5=2+3

mà 3>căn 2

nên 2+3>2+căn 2

=>5>2+căn 2

1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)

\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)

\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)

2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)

\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)

Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)

b) \(5=2+3=2+\sqrt{9}\)

Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)

3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

a) 1,2+3.1,3=5,1

b) 0,2+2.0,5=1,2

a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)

Bài 1: 

a) \(\sqrt{1.44}+3\sqrt{1.69}=1.2+3\cdot1.3=1.2+3.9=5.1\)

b) \(\sqrt{0.04}+2\cdot\sqrt{0.25}=0.2+2\cdot0.5=1.2\)

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

(100 số số hạng)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{100}{\sqrt{100}}=\dfrac{100}{10}=10\)

a: ĐKXĐ: x>=0; x<>4

\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\cdot\dfrac{\sqrt{x}-2+2}{2}\)

\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)

\(=\dfrac{2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)

TH1: M>=căn M

=>M^2>=M

=>M^2-M>=0

=>5*căn x-1>=0

=>x>=1/25 và x<>4

TH2: M<căn M

=>5căn x-1<0

=>x<1/25

Kết hợp ĐKXĐ, ta được: 0<=x<1/25