Ta có : ( 2x - 1 )²⁰²⁰ = ( 2x - 1 )²⁰²¹

=> ( 2x - 1 )²⁰²¹ - ( 2x - 1 )²⁰²⁰ = 0

=> ( 2x - 1 )²⁰²⁰ . [( 2x -1 )¹ - 1 ] = 0

=>  2x - 1 = 0               2x = 1                     x = 1/2

 hoặc                  =>                         =>

     2x - 1 = 1                2x = 2                     x =1

Vậy x = 1 hoặc x = 1/2

Số kẹo của chị là:

54x2= 108 (Viên kẹo)

ĐS:108 viên kẹo

Mong cậu ti-ck chi mik

108 viên kẹo

a) (x+3)² + (4+x)(4-x) = 10

x² + 6x + 9 + 16- x² = 10

6x + 25 = 10

6x = -15

x = -15/6

b) 9(x+1)² - (3x-2)(3x+2) = 10

9x² + 18x + 9 - 9x² + 4 =10

18x + 13 = 10

18x = -3

x = -1/6

a) ( x + 3 )² + ( 4 + x )( 4 - x ) = 10

⇔ x² + 6x + 9 + 16 - x² = 10

⇔ 6x + 25 = 10

⇔ 6x = -15

⇔ x = -15/6 = -5/2

b) 9( x + 1 )² - ( 3x - 2 )( 3x + 2 ) = 10

⇔ 9( x² + 2x + 1 ) - ( 9x² - 4 ) = 10

⇔ 9x² + 18x + 9 - 9x² + 4 = 10

⇔ 18x + 13 = 10

⇔ 18x = -3

⇔ x = -3/18 = -1/6

x² + xy + 5x + 5y = ( x² + xy ) + ( 5x + 5y ) = x( x + y ) + 5( x + y ) = ( x + y )( x + 5 )

x² - y² + 3x - 3y = ( x² - y² ) + ( 3x - 3y ) = ( x - y )( x + y ) + 3( x - y ) = ( x - y )( x + y + 3 )

x² + xy + 5x + 5y 

= (x²+ xy) + ( 5x+5y)

= x(x+y) + 5(x+y)

= (x+y)(x+5)

x² - y² + 3x - 3y

= (x² - y²) + ( 3x -3y)

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

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

chúc bạn học tốt ^^

\(A=\frac{x}{y}.\frac{x}{y^2}=\frac{x^2}{y^3}\left(\text{vì }x>0;y< 0\text{ nên: }\frac{x}{y^2}>0\right)\)

\(A=\frac{x}{y}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{x}{y}\cdot\frac{\sqrt{x^2}}{\sqrt{y^4}}=\frac{x}{y}\cdot\frac{\left|x\right|}{\left|y^2\right|}=\frac{x}{y}\cdot\frac{x}{y^2}=\frac{x^2}{y^3}\)( x > 0 ; y < 0 )

có toy đây

ko đăng những câu hỏi linh tinh nha bn

\(\sqrt{2006}-\sqrt{2005}=\frac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)   

\(\sqrt{2007}-\sqrt{2006}=\frac{\left(\sqrt{2007}-\sqrt{2006}\right)\left(\sqrt{2007}+\sqrt{2006}\right)}{\sqrt{2007}+\sqrt{2006}}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)   

Vì \(\sqrt{2006}+\sqrt{2005}< \sqrt{2007}+\sqrt{2006}\)   

Nên \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2006}}\)   

Vậy \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)

a) \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x\)

ĐKXĐ : x ≥ 0

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=-\left(x-4\right)\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=0\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}+x+2\right)=0\)

⇔ \(7\left(\sqrt{x}-2\right)=0\)

⇔ \(\sqrt{x}-2=0\)

⇔ \(\sqrt{x}=2\)

⇔ \(x=4\)( tm )

b) \(\frac{\sqrt{x}+5}{\sqrt{x}-4}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne16\end{cases}}\)

⇔ \(\left(\sqrt{x}+5\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)\)

⇔ \(x+8\sqrt{x}+15=x-6\sqrt{x}+8\)

⇔ \(x+8\sqrt{x}-x+6\sqrt{x}=8-15\)

⇔ \(14\sqrt{x}=-7\)

⇔ \(\sqrt{x}=-2\)( vô lí )

=> Phương trình vô nghiệm