Bài 1:

$x-1=|2x-1|\geq 0\Rightarrow x\geq 1$

$\Rightarrow 2x-1>0\Rightarrow |2x-1|=2x-1$. Khi đó:

$2x-1=x-1\Leftrightarrow x=0$  (không thỏa mãn vì $x\geq 1$)

Vậy không tồn tại $x$ thỏa đề.

Bài 2:

Nếu $x\geq \frac{1}{3}$ thì:

$3x-1=2x+3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{3}$ thì:

$1-3x=2x+3$

$\Leftrightarrow -2=5x\Leftrightarrow x=\frac{-2}{5}$ (tm)

Vậy......

Bài 3: Xét các TH sau:

TH1: $x\geq 2$ thì:

$x-1+x-2=3$

$2x-3=3$

$2x=6$

$x=3$ (thỏa mãn)

TH2: $1\leq x< 2$ thì:

$x-1+2-x=3$

$1=3$ (vô lý- loại)

TH3: $x< 1$

$1-x+2-x=3$

$3-2x=3$

$2x=0$

$x=0$ (thỏa mãn)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

a, 3x - 2x < 6 <=> x < 6 

b, đk : x khác -1 ; 3 

=> x^2 - 3x = x^2 - x - 2 

<=> -2x = -2 <=> x = 1 (tm) 

[x-1]/7=[-3]/[y+3]`

`=>(x-1)(y+3)=-21=-21.1=-1.21=-3.7=-7.3`

`@{(x-1=-21),(y+3=1):}=>{(x=-20),(y=-2):}`

`@{(x-1=-1),(y+3=21):}=>{(x=0),(y=18):}`

`@{(x-1=-3),(y+3=7):}=>{(x=-2),(y=4):}`

`@{(x-1=-7),(y+3=3):}=>{(x=-6),(y=0):}`

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)

Vì \(x\le3\Rightarrow\dfrac{2}{\sqrt{x}}\ge\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow-\dfrac{2}{\sqrt{x}}\le-\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{3}}\le1-\dfrac{2}{\sqrt{3}}\)

\(\Rightarrow\)\(P\le\dfrac{3-2\sqrt{3}}{3}\)

Dấu = xra khi x=3

Vậy \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

a: Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-x}+\dfrac{\sqrt{x}+3}{x+5\sqrt{x}+6}\)

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

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

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

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

\(\Leftrightarrow2x^3-3x^2+x+a=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow2\left(-27\right)-3\cdot9-3+a=0\\ \Leftrightarrow-54-27-3+a=0\\ \Leftrightarrow-84+a=0\\ \Leftrightarrow a=84\)