1.

$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$

$=x+3+(3-x)=6$

2.

$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

4.

$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$

$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$

\(A=\left(2x-3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(A=8x^3-12x^2+18x-12x^2+18x-27-8x^3+2\)

\(A=\left(8x^3-8x^3\right)+\left(-12x^2-12x^2\right)+\left(18x+18x\right)+\left(-27+2\right)\)

\(A=-24x^2+36x-25\)

\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)

\(=8x^3-8x^2+29\)

ta có 

\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)

\(A=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)

\(A=8x^3-8x^2+29\)

Trả lời:

A = ( 2x + 3 ) ( 4x² - 6x + 9 ) - 2 ( 4x² - 1 )

= 8x³ - 12x² + 18x + 12x² - 18x + 27 - 8x² + 2

= 8x³ - 8x² + 29

a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

mình chỉ biết làm một nửa k biết có đứng k bạn có chắc đề bài đúng k

5x^2 - 1^2 - (2x^3-3^3)= (5x^2-1x^2)-(2x^3-3^3) hdt số 3 và số 7

\(=8x^3-27-8x^3+50=23\)

Ta có: \(T=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)

\(=\left|x-3\right|+\left|x+3\right|\)

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

\(=6\)

12x\(^2\)+ 8x-9x-6-12x\(^2\)+30x+2x-5+1 = 31x - 10

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)

mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)

a,\(\dfrac{3-\sqrt{x}}{x-9}\)

=\(-\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

=\(-\dfrac{1}{3+\sqrt{x}}\)