A = |2x - 5| + 3 - 2x

A = 2x - 5 + 3 - 2x

A = (2x - 2x) + (-5 + 3)

A = -2

B = |x² - 5x + 4| - 4 + 5x - x²

B = x² - 5x + 4 - 4 + 5x - x²

B = (x² - x²) + (-5x + 5x) + (4 - 4)

B = 0

\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)

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

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

\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)

\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)

\(\Leftrightarrow7x^2-5x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)

Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)

`(3-2x)^2=(x-2)(2x-3)`

`<=>(2x-3)^2 -(x-2)(2x-3)=0`

`<=> (2x-3)(2x-3-x+2)=0`

`<=> (2x-3)(x-1)=0`

\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)

b) 3(x² -6x +9) - (16x² -1)   =   3x² -18x +27 -16x²+1 = -13x² -18x +28

c) sai đề rồi bạn mình sửa lại là 

(x+2)(x-2)-(x+2)²  = x² - 4 - (x²+4x+4) = 4x-8

d) (2x+3)(2x-3)-4(x+3)²  =  4x² - 9 - 4(x²+6x+9)  =  4x² -9 - 4x² - 24x - 36 = -24x -45

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

\(=\frac{2\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x-1\right)\left(x+1\right)}-\frac{2.2.\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

=\(=\frac{2x-2+4x^2-2x-4\left(x^2-1\right)}{2x\left(x-1\right)\left(x+1\right)}=\frac{2x-2+4x^2-2x-4x^2+4}{2x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{1}{x\left(x-1\right)\left(x+1\right)}\)

b,ta có

\(\frac{1}{P}=x\left(x-1\right)\left(x+1\right)\)

Vì x(x-1)(x+1) là 3 số liên tiếp

=>x(x-1)(x+1) chia hết cho 3

hay 1/p chia hết cho 3

M = (x - 3)³ - (x + 1)³ + 12x (x - 1)

= x³ - 9x² + 27x - 27 - (x³ + 3x² + 3x + 1) + 12x² - 12x

= x³ - 9x² + 27x - 27 - x³ - 3x² - 3x - 1 + 12x² - 12x

= (x³ - x³) + (12x² - 9x² - 3x²) + (27x - 3x - 12x) - (27 + 1)

= 12x - 28

\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8

d