a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)

\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)

b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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

a) \(A=\left(\sqrt{x}+3\right)^2-4\sqrt{x}-6\)

\(A=x+6\sqrt{x}+9-4\sqrt{x}-6\)

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

b) \(A=x+2\sqrt{x}-3\)

\(A=x+3\sqrt{x}-\sqrt{x}-3\)

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

\(A=\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)\)

a: A=x+6căn x+9-4căn x-6

=x+2căn x+3

b: A ko phân tích được nha bạn

d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)

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

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

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

\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)

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

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

\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)

\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(\text{f)}x-2\sqrt{x-1}-a^2\)

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

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

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

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

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

= (x + y - 3x)(x + y + 3x)

= (y - 2x)(4x + y)

e) (x + 5)² - (2x - 1)²

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

= (6 - x)(3x + 4)

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

(x - 10)(x + 2) = 0

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

mình tưởng lopw 9 mới học căn

em chịu bác lớp 7 học rui

lớp 7 đã học đâu?

a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)

\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)

\(=-7\sqrt{3}\cdot\sqrt{xy}\)

b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

$a)-7xy.\sqrt{\dfrac{3}{xy}}$

$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$

$=-7.\sqrt{\dfrac{xy}{3}}$

$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$

$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$

$=(\sqrt{a}+1)(b\sqrt{a}+1)$

a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)

b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)

b: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)