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

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

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

Bài 1:

a)\(Q=2x-\sqrt{x^2+2x+1}=2x-\sqrt{\left(x+1\right)^2}=2x-\left|x+1\right|\)

b)Tại x=7 thay vào Q ta được:

\(Q=2.7-\left|7+1\right|=14-8=6\)

Bài 2:

\(\sqrt{x^2-6x}+7x=13\)\(\Leftrightarrow\sqrt{x^2-6x}=13-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-7x\ge0\\x^2-6x=\left(13-7x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\0=48x^2-85x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\\Delta=\left(-85\right)^2-4.48.169=-25223< 0\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm.

a) Ta có:

\(A=2x+\sqrt{x^2-6x+9}\)

\(A=2x+\sqrt{\left(x-3\right)^2}\)

\(A=2x+\left|x-3\right|\)

Nếu \(x< 3\) thì: \(A=2x+3-x=x+3\)

Nếu \(x\ge3\) thì: \(A=2x+x-3=3x-3\)

b) Ta có: \(\left|x\right|=5\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Nếu x = 5: \(A=3\cdot5-3=12\)

Nếu x = -5: \(A=-5+3=-2\)

c) Ta có: \(A=2\Leftrightarrow\orbr{\begin{cases}x+3=2\\3x-3=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)

Vậy x = -1

a) \(A=2x+\sqrt{x^2-6x+9}\)

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

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

Với x ≥ 3 => A = 2x + x - 3 = 3x - 3

Với x < 3 => A = 2x + 3 - x = x + 3

b) | x | = 5 => x = ±5

Với x = 5 > 3 => A = 3.5 - 3 = 12

Với x = -5 < 3 => A = -5 + 3 = -2

c) A = 2 

⇔ 2x + | x - 3 | = 2

⇔ | x - 3 | = 2 - 2x (*)

Với x ≥ 3 

(*) ⇔ x - 3 = 2 - 2x

     ⇔ x + 3x = 2 + 3

     ⇔ 4x = 5

     ⇔ x = 5/4 ( ktm )

Với x < 3

(*) ⇔ 3 - x = 2 - 2x

     ⇔ -x + 2x = 2 - 3

     ⇔ x = -1 ( tm )

Vậy x = -1

c) Ta có: \(C=4\left(3x-2\right)^2+\left(4-x\right)^2-\left(6x-4\right)\left(8-2x\right)\)

\(=4\left(9x^2-12x+4\right)+x^2-8x+16-\left(48x-12x^2-32+8x\right)\)

\(=36x^2-48x+16+x^2-8x+16-48x+12x^2+32-8x\)

\(=49x^2-112x+64\)

\(=\left(7x-8\right)^2\)

\(=\left(7\cdot149-8\right)^2\)

\(=1071225\)

d) \(\left(3x-4\right)^2-9\left(x-2\right)\left(x+2\right)\)

\(=9x^2-24x+16-9\left(x^2-4\right)\)

\(=9x^2-24x+16-9x^2+36\)

\(=-24x+52\)

\(=-24\cdot\left(-2\right)+52\)

=48+52=100

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

\(=x\left(x^2-6x+9\right)-\left(x^2+4x-5\right)-x\left(x^2-4\right)\)

\(=x^3-6x^2+9x-x^2-4x+5-x^3+4x\)

\(=-7x^2+9x+5\)

\(=-7\cdot\left(-1\right)^2+9\cdot\left(-1\right)+5\)

\(=-7-9+5\)

=-16+5=-11

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

mn giúp e sớm đc ko ạ

Bài 2: 

\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Ta có: \(P=x^2-2x+2020\)

\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)

\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)

=2026

Bài 1: 

\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

=-6

Bài 1: (x-7)(x-8)-(x-5)(x-2)

=x^2 - 15x +56 -( x^2 -7x +10)

=46-8x.Thay x=-1/5 vào bt ta có:

A=46-8*(-1)/5=47,6

Bài 2:(x - 3)^2 - 2(x - 3)(x + 2)+ (x+2)^2

=(x - 3)[x - 3 - 2(x+2)] +(x+2)^2

=(x-3)[-x-7] + x^2+4x+4

=-x^2 -4x +21 +x^2+4x+4

=25

Bài 3:

a)2x^2 - 6x=0

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

<=>2x=0 hoặc x-3=0

<=>x=0 hoặc x=3

b)x^2-6x+9=0 <-- chắc đề thế này

<=>(x-3)^2=0 dùng HĐT

<=>x-3=0 =>x=3

Ôi! Mình cám ơn nhé <3