a) \(1=\sqrt{1}< \sqrt{2}\)

b) \(2=\sqrt{4}>\sqrt{3}\)

c) \(6=\sqrt{36}< \sqrt{41}\)

d) \(7=\sqrt{49}>\sqrt{47}\)

e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)

f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)

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

h) \(\sqrt{3}>0>-\sqrt{12}\)

i) \(5=\sqrt{25}< \sqrt{29}\)

\(\Rightarrow-5>-\sqrt{29}\)

a: \(\sqrt{a^2}=\left|a\right|\)

\(\sqrt[3]{a^3}=a\)

b: \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)

b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)

nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)

a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)

c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)

\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

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

`b,` Tớ tính mãi ko ra, xl cậu nha=')

Đặt \(A=\sqrt{a+1}+\sqrt{a+3}\)

\(\Rightarrow A^2=2a+4+2\sqrt{\left(a+1\right)\left(a+3\right)}\)

Đặt \(B=2\sqrt{a+2}\)

\(\Rightarrow B^2=4a+8\)

Xét hiệu \(B^2-A^2=2a+4-2\sqrt{\left(a+1\right)\left(a+3\right)}\)

Áp dụng BĐT Cô-si, ta có \(2a+4=\left(a+1\right)+\left(a+3\right)\) \(>2\sqrt{\left(a+1\right)\left(a+3\right)}\) 

 (Dấu "=" không thể xảy ra vì khi đó sẽ suy ra đẳng thức vô lí là \(1=3\))

 Từ đó suy ra \(B^2-A^2>0\) \(\Leftrightarrow B^2>A^2\), và do A, B dương nên suy ra \(B>A\). Nói cách khác, \(2\sqrt{a+2}>\sqrt{a+1}+\sqrt{a+3}\)

sao em bấm máy tính thì dấu bằng $x$ảy ra khi $x$=10000000 vậy ạ.

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)

1) c/m \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

áp dụng BĐT cô shi cho 2 số thực dương ta có:

\(a+b\ge2\sqrt{ab}\);\(b+c\ge2\sqrt{bc}\);\(a+c\ge2\sqrt{ac}\)

cộng vế vs vế:\(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

↔\(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

dấu = xảy ra khi a=b=c

vậy...

b)ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{25}}\)→\(A>\frac{1}{\sqrt{25}}+\frac{1}{\sqrt{25}}+...+\frac{1}{\sqrt{25}}\)(25 số hạng)

\(A>\frac{25}{\sqrt{25}}=\sqrt{25}=5\)

vậy.....