Câu 2:

a:

\(10=2\cdot5;12=2^2\cdot3;18=3^2\cdot2\)

=>\(BCNN\left(10;12;18\right)=3^2\cdot2^2\cdot5=180\)

 \(x⋮10;x⋮12;x⋮18\)

=>\(x\in BC\left(10;12;18\right)\)

=>\(x\in B\left(180\right)\)

=>\(x\in\left\{180;360;540;...\right\}\)

mà 100<x<500

nên \(x\in\left\{180;360\right\}\)

b:

\(72=2^3\cdot3^2;24=2^3\cdot3;120=2^3\cdot3\cdot5\)

=>\(ƯCLN\left(72;24;120\right)=2^3\cdot3=24\)

 \(72⋮x;24⋮x;120⋮x\)

=>\(x\inƯC\left(72;24;120\right)\)

=>\(x\inƯ\left(24\right)\)

=>\(x\in\left\{1;2;3;4;6;8;12;24\right\}\)

mà 5<x<10

nên \(x\in\left\{6;8\right\}\)

Bài 1.1 

a. Để căn thức có nghĩa (CTCN) thì $2x-1\geq 0$

$\Leftrightarrow x\geq \frac{1}{2}$

b. Để CTCN thì $-2x+0,5\geq 0$

$\Leftrightarrow 0,5\geq 2x\Leftrightarrow x\leq \frac{1}{4}$

c. Để CTCN thì \(\left\{\begin{matrix} x-1\neq 0\\ \frac{1}{x-1}\geq 0\end{matrix}\right.\Leftrightarrow x-1>0\Leftrightarrow x>1\)

d. Để CTCN thì \(\left\{\begin{matrix} x^2+2021\neq 0\\ \frac{2022-x}{x^2+2021}\geq 0\end{matrix}\right.\Leftrightarrow 2022-x\geq 0\) (do $x^2+2021>0$ với mọi $x\in\mathbb{R}$)

$\Leftrightarrow x\leq 2022$

Bài 1.2

a. $3=\sqrt{9}>\sqrt{8}$
b. $-7=-\sqrt{49}> -\sqrt{51}$

c. $3+\sqrt{2}> 3+\sqrt{1}=4=2+2=2+\sqrt{4}> 2+\sqrt{3}$

d. $\sqrt{26}+3>\sqrt{25}+3=8=\sqrt{64}>\sqrt{63}$

e.

$\frac{1}{2}=\frac{2-1}{2}=\frac{\sqrt{4}-1}{2}> \frac{\sqrt{2}-1}{2}$

f.

Xét hiệu $5-2\sqrt{7}-(3-\sqrt{10})=2-(\sqrt{28}-\sqrt{10})$

$=2-\frac{18}{\sqrt{28}+\sqrt{10}}< 2-\frac{18}{\sqrt{2(28+10)}}$ (áp dụng BĐT $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$)

$=2-\frac{18}{\sqrt{76}}< 2-\frac{18}{\sqrt{81}}=0$

$\Rightarrow 5-2\sqrt{7}< 3-\sqrt{10}$

Bài 2.

a.

$\sqrt{(1-\sqrt{2})^2}=|1-\sqrt{2}|=\sqrt{2}-1$

b.

$=|\sqrt{5}-\sqrt{2}|+|\sqrt{5}+\sqrt{2}|$

$=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}$

c.

$C=\sqrt{7+\sqrt{3}}-\sqrt{7-\sqrt{3}}$

$C^2=7+\sqrt{3}+7-\sqrt{3}-2\sqrt{(7+\sqrt{3})(7-\sqrt{3})}$

$=14-2\sqrt{46}$

$\Rightarrow C=\sqrt{14-2\sqrt{46}}$ (do $C>0$)

c.

$\sqrt{5+2\sqrt{6}}=\sqrt{2+2\sqrt{2}.\sqrt{3}+3}$

$=\sqrt{(\sqrt{2}+\sqrt{3})^2}=|\sqrt{2}+\sqrt{3}|=\sqrt{2}+\sqrt{3}$

d.

$\sqrt{9-4\sqrt{5}}=\sqrt{2^2-2.2\sqrt{5}+5}$
$=\sqrt{(2-\sqrt{5})^2}=|2-\sqrt{5}|=\sqrt{5}-2$
f.

\(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2}=\frac{\sqrt{3-2\sqrt{3}+1}}{2}=\frac{\sqrt{(\sqrt{3}-1)^2}}{2}=\frac{|\sqrt{3}-1|}{2}=\frac{\sqrt{3}-1}{2}\)

2:

a: =(x-y)^2-4

=(x-y-2)(x-y+2)

b: =49-(16x^2-8xy+y^2)

=49-(4x-y)^2

=(7-4x+y)(7+4x-y)

3:

a: =x^2(x^4-x^2+2x+2)

b: =(x+y-x+y)[(x+y)^2+(x-y)(x+y)+(x-y)^2]

=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)

=2y(3x^2+y^2)

11 c)

\(a^2+2\ge2\sqrt{a^2+1}\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) (luôn đúng)

12 a)  Có a+b+c=1\(\Rightarrow\) (1-a)(1-b)(1-c)= (b+c)(a+c)(a+b) (*)

áp dụng BĐT cô-si: \(\left(b+c\right)\left(a+c\right)\left(a+b\right)\ge2\sqrt{bc}2\sqrt{ac}2\sqrt{ab}=8\sqrt{\left(abc\right)2}=8abc\) ( luôn đúng với mọi a,b,c ko âm ) 

b)  áp dụng BĐT cô-si: \(c\left(a+b\right)\le\dfrac{\left(a+b+c\right)^2}{4}=\dfrac{1}{4}\)

Tương tự: \(a\left(b+c\right)\le\dfrac{1}{4};b\left(c+a\right)\le\dfrac{1}{4}\)

\(\Rightarrow abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{4}=\dfrac{1}{64}\)

13 b) \(\left(a+b\right)\left(ab+1\right)\ge2\sqrt{ab}.2\sqrt{ab}=4ab\)

Dấu = xảy ra khi a=b=1