a) 2.I3x - 1I + 1 = 5
<=>2.I3x - 1I = 5-1
<=>2.I3x - 1I =4
<=>I3x - 1I=2
=>Có 2 trường hợp
3x-1=2 =>3x=3 =>x=1
3x-1=-2 =>3x=1 =>x=1/3
Vậy x có 2 giá trị thỏa mãn là 1 và 1/3

Hok tốt 

Bài 1 :

a) \(x^3-x^2-x-2=0\)

\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2=0\)

\(\Leftrightarrow\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)=0\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)=0\)(1)

Vì \(x^2+x+1=x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow x^2+x+1\ge\frac{3}{4}\forall x\)(2)

Từ (1) và (2) \(\Rightarrow x-2=0\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Bài 2: 

\(2x^2+y^2-2xy+2y-6x+5=0\)

\(\Leftrightarrow x^2-2xy+y^2-2x+2y+1+x^2-4x+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2-2\left(x-y\right)+1+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-y-1\right)^2+\left(x-2\right)^2=0\)(1)

Vì \(\left(x-y-1\right)^2\ge0\forall x,y\); \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2\ge0\forall x,y\)(2)

Từ (1) và (2) \(\Rightarrow\left(x-y-1\right)^2+\left(x-y\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x-1\\x=2\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=2\end{cases}}\)

Vậy \(x=2\)và \(y=1\)

\(=\frac{x-2\sqrt{x}+y+2\sqrt{y}-2\sqrt{xy}+1}{x-2\sqrt{xy}+y-1}\)\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1}{\left(\sqrt{x}-\sqrt{y}\right)^2-1}\)

\(=\frac{\left(\sqrt{x}-\sqrt{y}-1\right)^2}{\left(\sqrt{x}-\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)}=\frac{\sqrt{x}-\sqrt{y}-1}{\sqrt{x}-\sqrt{y}+1}\)

hok tốt !

6³  

HT 

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Đặt A =\(\frac{2x+1}{\sqrt{x}}\)( ĐKXĐ : \(x>0\))

Có : A = \(\frac{2x+1}{\sqrt{x}}=\frac{2x}{\sqrt{x}}+\frac{1}{\sqrt{x}}=2\sqrt{x}+\frac{1}{\sqrt{x}}\)

Áp dụng bđt AM-GM cho 2 số \(2\sqrt{x}\)và \(\frac{1}{\sqrt{x}}\)có :

\(2\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\sqrt{2}\)

\(\Rightarrow\)\(\frac{2x+1}{\sqrt{x}}\ge2\sqrt{2}\)

Vậy biểu thức có gtnn là \(\frac{2x+1}{\sqrt{x}}=2\sqrt{2}\Leftrightarrow2\sqrt{x}=\frac{1}{\sqrt{x}}\)

                                                                           \(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

Đề có sai ko ạ, theo mình nên thêm -3x vào P(x), vì tính ra số vô tỉ

\(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

= \(\sqrt{5-2\sqrt{2+\sqrt{8+4\sqrt{2}+1}}}\)

= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)

= \(\sqrt{5-2\sqrt{2+2\sqrt{2}+1}}\)

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

= \(\sqrt{5-2\left(\sqrt{2}+1\right)}\)

= \(\sqrt{5-2\sqrt{2}-2}\)

= \(\sqrt{\left(\sqrt{2}-1\right)^2}\)

= \(\sqrt{2}-1\)

CẢM ƠN BẠN RẤT NHIỀU <3

Câu 2 : \(f\left(x\right)=x^3-ax^2+bx-a\)

Áp dụng định lý Bezout ta có:

\(f\left(x\right)⋮\left(x-1\right)\)\(\Rightarrow f\left(1\right)=0\)

\(\Rightarrow1^3-a.1^2+b.1-a=1-a+b-a=0\)

\(\Leftrightarrow1-2a+b=0\)\(\Leftrightarrow2a-b=1\)(1)

\(\Rightarrow3\left(2a-b\right)=3\)\(\Rightarrow6a-3b=3\)(2)

\(f\left(x\right)⋮\left(x-3\right)\)\(\Rightarrow f\left(3\right)=0\)

\(\Rightarrow3^3-a.3^2+3b-a=27-9a+3b-a=0\)

\(\Leftrightarrow27-10a+3b=0\)\(\Leftrightarrow10a-3b=27\)(3)

Từ (2) và (3)

\(\Rightarrow\left(10a-3b\right)-\left(6a-3b\right)=27-3\)

\(\Leftrightarrow10a-3b-6a+3b=24\)

\(\Leftrightarrow4a=24\)\(\Leftrightarrow a=6\)

Thay \(a=6\)vào (1) ta có:

\(2.6-b=1\)\(\Leftrightarrow12-b=1\)\(\Leftrightarrow b=11\)

Vậy \(a=6\)và \(b=11\)