bài nào cũng đc ạ:33
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x\in\left\{6;8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$)
$=2-\frac{18}{\sqrt{76}}< 2-\frac{18}{\sqrt{81}}=0$
$\Rightarrow 5-2\sqrt{7}< 3-\sqrt{10}$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(a\left(b+c\right)\le\dfrac{1}{4};b\left(c+a\right)\le\dfrac{1}{4}\)
\(\Rightarrow abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{4}=\dfrac{1}{64}\)