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a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{x-3\sqrt{x}}\right):\dfrac{2}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{2}\)

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

b) Thay \(x=3-2\sqrt{2}\) vào A, ta được:

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

c) Để \(A< \dfrac{2}{3}\) thì \(\dfrac{\sqrt{x}+1}{2\sqrt{x}}-\dfrac{2}{3}< 0\)

\(\Leftrightarrow\dfrac{3\left(\sqrt{x}+1\right)-4\sqrt{x}}{6\sqrt{x}}< 0\)

\(\Leftrightarrow-\sqrt{x}+3< 0\)

\(\Leftrightarrow-\sqrt{x}< -3\)

\(\Leftrightarrow\sqrt{x}>3\)

hay x>9

Vậy: Để \(A< \dfrac{2}{3}\) thì x>9

a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)

\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)

\(=-16x+8\)

b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

=27x-55

4) (3x-2)(x-3)= 3x(x-3)-2(x-3)

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

=\(3x^2\)-9x-4x+6

=\(3x^2\)+(-9x-4x)+6

=\(3x^2\)-13x+6

5) (2x+1)(x+3)=2x(x+3)+1(x+3)

=2x.x+2x.3+1.x+1.3

=\(2x^2\)+6x+1x+3

=\(2x^2\)+(6x+1x)+3

=\(2x^2\)+7x+3

6) (x-3)(3x-1)=x(3x-1)-3(3x-1)

=x.3x+x.(-1)-3.3x-3.(-1)

=\(3x^2\)-1x-9x+3

=\(3x^2\)+(-1x-9x)+3

=\(3x^2\)-10x+3

rút gọn biểu thức

A) \(x^2\)-(x+4)(x-1)=\(x^2\)- x(x-1)-4(x-1)

=\(x^2\)-x.x-x.(-1)-4.x-4.(-1)

=\(x^2\)-\(x^2\)+1x-4x+4

=(\(x^2-x^2\))+(1x-4x)+4

= -3x+4

B) x(x+2)-(x-2)(x+4)=x.x+x.2-x(x+4)+2(x+4)

=\(x^2+2x\)-x.x-x.4+2.x+2.4

=\(x^2+2x-x^2-4x+2x+8\)

=(\(x^2-x^2\))+(2x-4x+2x)+8

=8

tính giá trị biểu thức

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

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

=3x-6-2.x-2.(-3)-x.x-x(-3)

=3x-6-2x+6-\(x^2\)+3x

=(3x-2x+3x)+(-6+6)\(-x^2\)

=4x - \(x^2\)

thay x=-8 vào biểu thức thu gọn ta được:

4.(-8)- (-8)\(^2\)

= - 32 +64

= 32

B= x(3-x)-(1+x)(1-x)

=x.3+x.(-x)-1(1-x)-x(1-x)

=3x -\(x^2\)-1.1-1 .(-x)-x.1-x.(-x)

=3x\(-x^2\)-\(1^2\)+1x-1x+\(x^2\)

=(3x+1x-1x)+(\(-x^2+x^2\))-1

=3x-1

thay x=-5 vào biểu thức thu gọn ta được:

3.(-5)-1

=-15-1

=-16

Thu gọn biểu thức

4) (3x - 2) (x - 3) 

= ( 3x² - 2x ) - ( 3x x 3 - 2 x 3 )

= 3x² - 2x - 3x x 3 + 2 x 3

= 3x² - 2x - 9x + 6

= 3x² - 11x + 6 

5) (2x + 1) (x + 3) 

= ( 2x² + 1x ) + ( 6x + 3 )

= 2x² + 1x + 6x + 3

= 2x² + 7x + 3

6) (x - 3) (3x - 1) 

= ( 3x² - 9x ) - ( x - 3 )

= 3x² - 9x - x + 3

= 3x² - 10 + 3

Rút gọn biểu thức

A) x^2 - (x + 4) (x - 1)

= x² - ( x² + 4x ) - ( x + 4 )

= x² - x² - 4x - x - 4

= -5x - 4

B) x (x + 2) - (x - 2) (x + 4)

= x² + 2x - ( x² - 2x ) + ( 4x - 8 )

= x² + 2x - x² + 2x + 4x - 8

= 8x - 8

Tính giá trị biểu thức

A = 3 (x - 2) - (2 + x) (x - 3) tại x = - 8

Thế x = -8 vào, ta có :

= 3 ( -8 -2 ) - ( 2 + -8 ) ( -8 - 3 )

= 3 x ( -10 ) - ( - 6 ) ( -11 )

= -30 - 66

= -96

B = x (3 - x) - (1 + x) ( 1 - x) tại x = - 5

Thế x = - 5 vào, ta có :

= -5 ( 3 - -5 ) - ( 1+ -5 ) ( 1 - -5 )

= -5 x 8 - (-4) x 6

= - 40 - -24

= -40 + 24

= -16

100% đúng 

hok tốt nha 

B1:

a,\(\left(3x-2\right)\left(x-3\right)=3x^2-9x-2x+6=3x^2-11x+6\)

b,\(\left(2x+1\right)\left(x+3\right)=2x^2+6x+x+3=2x^2+7x+3\)

c,\(\left(x-3\right)\left(3x-1\right)=3x^2-x-9x+3=3x^2-10x+3\)

B2:

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

2)\(x\left(x+2\right)-\left(x-2\right)\left(x+4\right)=x^2+2x-\left(x^2+4x-2x-8\right)\)

\(=x^2+2x-x^2-4x+2x+8=8\)

Bài 8:

\(F=x^2-2x+1+x^2-6x+9=2x^2-8x+10\\ F=2\left(x^2-4x+4\right)+2=2\left(x-2\right)^2+2\ge2\\ F_{min}=2\Leftrightarrow x=2\)

Bài 9:

\(A=-x^2+2x-1+5=-\left(x-1\right)^2+5\le5\\ A_{max}=5\Leftrightarrow x=1\\ B=-x^2+10x-25+2=-\left(x-5\right)^2+2\le2\\ B_{max}=2\Leftrightarrow x=5\\ C=-x^2+6x-9+9=-\left(x-3\right)^2+9\le9\\ C_{max}=9\Leftrightarrow x=3\)

`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

a) \(\left(2a-3\right)\left(a+1\right)-\left(a^2+6a+9\right):\left(a+3\right)\)

\(=\left(2a^2+2a-3a-3\right)-\left(a+3\right)^2:\left(a+3\right)\)

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

\(=2a^2-a-3-a-3\)

\(=2a^2-2a-6\)

b) \(\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\)

\(=\left(3x-5y\right)\cdot x^2y^2-3x^2y^2+4x^2y^3\)

\(=3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\)

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

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

\(=x\left(x^2-4x+4\right)-\left(x^3+8\right)+4x^2\)

\(=x^3-4x^2+4x-x^3-8+4x^2\)

\(=\left(x^3-x^3\right)+\left(-4x^2+4x^2\right)+4x-8\)

\(=4x-8\)