A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)

\(=8x^4y^2+4xy^4-28x^2y^3\)

B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)

\(=10x^4-5x^2-5x-4x^3+2x+2\)

\(=10x^4-5x^3-3x-4x^3+2\)

C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)

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

D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)

( Bài này áp dụng hằng đẳng thức là làm được ạ )

9(a + b)² - (a + b) = (a + b)[9(a + b) - 1]

(mx + my) + (3x + 3y) = m(x + y) + 3(x + y) = (m + 3)(x + y)

(12xy) - 6x - (2y - 1) = 6x(2y - 1) - (2y - 1) = (6x - 1)(2y - 1)

(7xy² - 5x²y) + (5x - 7y) = xy(7y - 5x) + (5x - 7y) = -xy(5x - 7y) + (5x - 7y) = (-xy + 1)(5x - 7y)

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

a) 9( a + b )² - ( a + b ) = ( a + b )[ 9( a + b ) - 1 ]

b) ( mx + my ) + ( 3x + 3y ) = m( x + y ) + 3( x + y ) = ( m + 3 )( x + y )

c) 12xy - 6x - ( 2y - 1 ) = 6x( 2y - 1 ) - ( 2y - 1 ) = ( 6x - 1 )( 2y - 1 )

d) ( 7xy² - 5x²y ) + ( 5x - 7y ) = xy( 7y - 5x ) + ( 5x - 7y ) = -xy( 5x - 7y ) + ( 5x - 7y ) = ( -xy + 1 )( 5x - 7y )

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

a) x²(5x³ – x - \(\dfrac{1}{2}\) )= x². 5x³ + x² . (-x) + x² . (-\(\dfrac{1}{2}\))

= 5x⁵ – x³ – \(\dfrac{1}{2}\)x²

b) (3xy – x² + y)\(\dfrac{2}{3}\)x²y = \(\dfrac{2}{3}\)x²y . 3xy + \(\dfrac{2}{3}\)x²y . (- x²) + \(\dfrac{2}{3}\)x²y . y

= 2x³y² – \(\dfrac{2}{3}\)x⁴y + \(\dfrac{2}{3}\)x²y²

c) (4x³– 5xy + 2x)(- \(\dfrac{1}{2}\)xy) = - \(\dfrac{1}{2}\)xy . 4x³ + (- \(\dfrac{1}{2}\)xy) . (-5xy) + (- \(\dfrac{1}{2}\)xy) . 2x

= -2x⁴y + \(\dfrac{5}{2}\)x²y² - x²y.

a) x² (5x³ - x - \(\dfrac{1}{2}\))

= 5x⁵ - x³ - \(\dfrac{1}{2}\)x²

b) (3xy - x² + y) \(\dfrac{2}{3}\)x²y

= 2x³y² - \(\dfrac{2}{3}\)x⁴y + \(\dfrac{2}{3}\)x²y²

c) (4x³ - 5xy +2x) (-\(\dfrac{1}{2}\)xy)

= -2x⁴y + \(\dfrac{5}{2}\)x²y² - x²y

a. ĐKXĐ: ..

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)

b.

Thế pt trên xuống dưới:

\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)

\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)

\(\Leftrightarrow xy\left(2x^2-2xy-y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)

Thế vào pt đầu ...

Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều

Thực hiện phép tính:

³.4xy²+y².4xy²-7xy.4xy²

^=>8x⁴y²+4xy⁴-28x²y³

c: =(2x^2-3)[(2x^2)^2+2x^2*3+3^2]

=8x^6-27

d:\(=\left(3x^2-2y-2x^2+y\right)\left(9x^4-12x^2y+4y^2+6x^4-3x^2y-4x^2y+2y^2+4x^4-4x^2y+y^2\right)\)

\(=\left(x^2-y\right)\left(19x^4-23x^2y+7y^2\right)\)

Làm tính nhân

(4x³+3xy²-2y³).(3x²-5xy-6y²)

=12x⁵+12y⁵-20x⁴y-36x²y³-8xy⁴

Phân tích đa thức thành nhân tử

10x³+5x²y-10x²y-10xy²+5y³

=10x³-5x²y-10xy²+5y³

=5(2x³-x²y-2xy²+y³-)

Bài 2 :

a) \(\left(5x^2y-8xy^2+y^3\right)\left(2x^3+x^2y-3y^2\right)\)

\(=10x^5y+5x^4y^2-15x^2y^3-16x^4y^2-8x^3y^3+24xy^4+2x^3y^3+x^2y^4-3y^5\)

\(=10x^5y-11x^4y^2-6x^3y^3+x^2y^4-15x^2y^3+24xy^4-3y^5\)

Đề có vẻ sai sai. Bạn xem lại đề xem có đúng không?