a, 3x - 2x < 6 <=> x < 6 

b, đk : x khác -1 ; 3 

=> x^2 - 3x = x^2 - x - 2 

<=> -2x = -2 <=> x = 1 (tm) 

Giải pt à bạn

Bài 1:

a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)

\(=x^2+x+x^2-x-2x^2\)

\(=2x^2-2x^2\)

\(=0\)

b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)

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

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

\(=2x^2+4\)

c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)

\(=\left(x-3+x+7\right)^2\)

\(=\left(2x+4\right)^2\)

Bài 2:

a) \(x^2-4-4\left(2-x\right)\)

\(=\left(x^2-4\right)+4\left(x-2\right)\)

\(=\left(x^2-2^2\right)+4\left(x-2\right)\)

\(=\left(x+2\right)\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2+4\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

b) \(x^3+8\left(1-2x\right)^3\)

\(=x^3+\left[2\left(1-2x\right)\right]^3\)

\(=\left[x+2\left(1-2x\right)\right]\left[x^2-x\cdot2\left(1-2x\right)+2^2\left(1-2x\right)^2\right]\)

\(=\left(x+2-4x\right)\left[x^2-2x\left(1-2x\right)+4\left(1-4x+4x^2\right)\right]\)

\(=\left(2-3x\right)\left(x^2-2x+4x^2+4-16x+16x^2\right)\)

\(=\left(2-3x\right)\left(21x^2-18x+4\right)\)

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

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 \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)

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)

\(\left(2x-1\right):\dfrac{10}{7}=\dfrac{28}{15}:\dfrac{4}{3}\)

\(\Rightarrow\left(2x-1\right).\dfrac{7}{10}=\dfrac{28}{15}.\dfrac{3}{4}\)

\(\Rightarrow\left(2x-1\right).\dfrac{7}{10}=\dfrac{7}{5}\)

\(\Rightarrow2x-1=\dfrac{7}{5}:\dfrac{7}{10}\)

\(\Rightarrow2x-1=\dfrac{7}{5}.\dfrac{10}{7}\)

\(\Rightarrow2x-1=2\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\dfrac{3}{2}\)

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 \(y=\dfrac{1}{3}\)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

\(\left(x-3\right)\cdot\left(y-5\right)=3\)

=>\(\left(x-3\right)\cdot\left(y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)

=>\(\left(x-3;y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(4;8\right);\left(6;6\right);\left(2;2\right);\left(0;4\right)\right\}\)