ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)

a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

\(=\dfrac{x.\left(3x-y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{x.\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{2xy}{9x^2-y^2}\)

\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)

\(=\dfrac{2x}{3x-y}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)

\(=\dfrac{x+5}{x.\left(x+5\right)}\)

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

a) Ta có: \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

\(=\dfrac{x\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{x\left(3x+y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{2xy}{\left(3x+y\right)\left(3x-y\right)}\)

\(=\dfrac{3x^2-xy+3x^2+xy+2xy}{\left(3x+y\right)\left(3x-y\right)}\)

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

\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right)\left(3x-y\right)}\)

\(=\dfrac{2x}{3x-y}\)

b) Ta có: \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x\left(x+5\right)}-\dfrac{3x}{x\left(x+5\right)}\)

\(=\dfrac{4x+5-3x}{x\left(x+5\right)}\)

\(=\dfrac{x+5}{x\left(x+5\right)}\)

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

\(a,\left(2x-y\right)\left(2x+y\right)-\left(2x+y\right)^2\)

\(=4x^2-y^2-4x^2-4xy+y^2\)

\(=-4xy\)

\(b,\left(x-3\right)\left(x^2+3x+9\right)-\left(5-x\right)^2\)

\(=x^3-27-25+10x-x^2\)

\(=x^3-x^2+10x-52\)

\(c,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x+y\right)^3\)

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

\(d,\left(3x-5\right)^2-\left(3x+5\right)^2\)

\(=\left(3x-5-3x-5\right)\left(3x-5+3x+5\right)\)

\(=-10.6x=-60x\)

a) ( x + 2 )( x² - 2x + 4 ) - ( 18 + x³ )

= x³ + 8 - 18 - x³ = -10

b) ( 2x - y )( 4x² + 2xy + y² ) - ( 2x + y )( 4x² - 2xy + y² )

= 8x³ - y³ - ( 8x³ + y³ )

= 8x³ - y³ - 8x³ - y³ = -2y³

c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )

= x² - 9 - ( x² + 4x - 5 )

= x² - 9 - x² - 4x + 5 = -4x - 4

d) ( 3x - 2 )² + ( x + 1 )² + 2( 3x - 2 )( x + 1 )

= ( 3x - 2 + x + 1 )² 

= ( 4x - 1 )² 

a) (x+3)(x^2-3x+9)-(54+x^3)

= x^3- 3x^2+9x+3x^2-9x+27-54-x63

= -27

b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)

= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]

= [(2x)3^3+ y^3] – [(2x)^3 – y^3]

= (2x)^3 + y^3 – (2x)^3 + y^3

= 2y^3

a)(x+3)(X^2-3x+9)-(54+x^3)

= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)

= 27- 54

= -27

b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)

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

= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)

= \(2y^3\)

a) Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+27-54-x^3\)

=-27

bạn ghi lại đề được không vậy? đọc không hiểu gì hết :))

Trả lời:

a, ( 2x + y )³ - ( 2x - y )³ 

= ( 2x )³ + 3.( 2x )².y + 3.2x.y² + y³ - [ ( 2x )³ - 3.( 2x )².y + 3.2x.y² - y³ ] 

= 8x³ + 12x²y + 6xy² + y³ - ( 8x³ - 12x²y + 6xy² - y³ )

= 8x³ + 12x²y + 6xy² + y³ - 8x³ + 12x²y - 6xy² + y³

= 24x²y + 2y³

b, ( 5 - 3x )³ - ( 5 + 3x )³ 

= 5³ - 3.5².3x + 3.5.( 3x )² - ( 3x )³ - [ 5³ + 3.5².3x + 3.5.( 3x )² + ( 3x )³ ]

= 125 - 225x + 135x² - 27x³ - ( 125 + 225x + 135x² + 27x³ )

= 125 - 225x + 135x² - 27x³ - 125 - 225x - 135x² - 27x³

=  - 54x³- 450x​

c, ( x + 3 ) ( x² - 2x + 9 ) - ( 54 + x³ )

= x³ - 2x² + 9x + 3x² - 6x + 27 - 54 - x³

= x² + 3x - 27

d, ( 2x + y ) ( 4x² - 2xy + y² ) - 2xy ( 4x² + 2xy + y² )

= ( 2x )³ + y³ - 8x³y - 4x²y² - 2xy³ 

= 8x³ + y³ - 8x³y - 4x²y² - 2xy³

a) (2x-y)(2x+y)-(2x+y)^2 

= 4x²-y²-(4x²+4xy+y²)

=  4x²-y²-4x²-4xy-y²

^{= -4xy}

b) (x-3)(x^2+3x+9)-(5-x)^2 

= (x³-27)-(25-10x+x²)

= x³-27-25+10x-x²

= x³-x²+10x-52

c) (2x+y)(4x^2-2xy+y^2)-(2x+y)^3   

= (2x)³+y³- ((2x)³+3.4x².y+3.y².2x+y³)

= 8x³+y³-(8x³+12x²y+6xy²+y³)

= 8x³+y³-(8x³+12x²y+6xy²+y³)

= 8x³+y³-8x³-12x²y-6xy²-y³

⁼-12x²y-6xy²

 d) (3x-5)^2-(3x+5)^2

= (3x-5-3x-5)(3x-5+3x+5)

= -10.6x

= -60x 

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)