Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

a) (x+9)(x-9)-x²=x²-81-x²=-81

b) (10x-1)(10x+1)-(10x-1)²=100x²-1-100x²+20x-1=20x-2

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

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

\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)

a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

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

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

\(=\left[\left(x+3\right)+\left(x-3\right)\right]^2\)

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

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

\(=4x^2\)

b) \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=\left(64x^3-48x^2+12x-1\right)-\left(64x^3+12x-48x^2-9\right)\)

\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)

\(=\left(64x^3-64x^3\right)-\left(48x^2-48x^2\right)+\left(12x-12x\right)-\left(1-9\right)\)

\(=0-0+0+8\)

\(=8\)

a) (x + 3)² + (x - 3)² + 2(x² - 9)

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

= (x + 3 + x - 3)²

= (2x)²

= 4x²

b) (4x - 1)³ - (4x - 3)(16x² + 3)

= 64x³ - 48x² + 12x - 1 - 64x³ - 12x + 48x² + 9

= (64x³ - 64x³) + (-48x² + 48x²) + (12x - 12x) + (-1 + 9)

= 8

Bài 1:

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)

\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)

Bài 2:

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

b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)

`a)(x-1)^2-(x-2)(x+2)`

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

`=-2x+5`

`b)(2x+4)(8x-3)(4x+1)^2`

`=(16x^2-6x+32x-12)(16x^2+8x+1)`

`=(16x^2-26x-12)(16x^2+8x+1)`

`=256x^4+128x^3+16x^2-416x^3-208x^2-26x-192x^2-96x-12`

`=256x^4-288x^3-384x^2-122x-12`

`c)(a+2)^3-a(a-3)^2`

`=a^3+6a^2+12a+8-a(a^2-6a+9)`

`=a^3+6a^2+12a+8-a^3+6a^2-9a`

`=12a^2+3a+8`

\(A=5x^2-3x-x^3+x^2+x^3-62x-10+3x\\ A=6x^2-62x-10\\ B=x^3+x^2+x-x^3-x^2-x+5=5\\ C=3x^2y-15xy^2+15xy^2-10y^3+10y^2-3x^2y-4=-4\)

b: Ta có: \(B=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

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

=5

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne9\right)\)

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

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

\(=\dfrac{x+3\sqrt{x}}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{7-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{7-\sqrt{x}}=\dfrac{x}{\sqrt{x}-7}\)

\(B=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\left(x>0,x\ne1\right)\)

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

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

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

3:

a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

2:

a: 2căn 7=căn 28

3căn 2=căn 18

mà 28>18

nên 2*căn 7>3*căn 2

b: 5=2+3

mà 3>căn 2

nên 2+3>2+căn 2

=>5>2+căn 2

1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)

\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)

\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)

2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)

\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)

Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)

b) \(5=2+3=2+\sqrt{9}\)

Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)

3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)