có cậu đấy Đỗ Thùy Dương

Đỗ Thuỳ Dương có cần mik cho bn một vé báo cáo ko

,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l  = 3/4
Hoặc 3x-1 = 3/4
 <=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1

a) ( 5x - y )( 25x² + 5xy + y² ) = ( 5x )³ - y³ = 125x³ - y³

b) ( x - 3 )( x² + 3x + 9 ) - ( 54 + x³ ) = x³ - 3³ - 54 - x³ = -27 - 54 = -81

c) ( 2x + y )( 4x² - 2xy + y² ) - ( 2x - y )( 4x² + 2xy + y² ) = ( 2x )³ + y³ - [ ( 2x )³ - y³ ]= 8x³ + y³ - 8x³ + y³ = 2y³

d) ( x + y )² + ( x - y )² + ( x + y )( x - y ) - 3x² = x² + 2xy + y² + x² - 2xy + y² + x² - y² - 3x² = y²

e) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )²

= x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 )

= x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6

= -3x² + 39x + 6

= -3( x² - 13x - 2 )

f) ( x + y )( x² - xy + y² ) + ( x - y )( x² + xy + y² ) - 2x³

= x³ + y³ + x³ - y³ - 2x³

= 0

g) x² + 2x( y + 1 ) + y² + 2y + 1

= x² + 2x( y + 1 ) + ( y² + 2y + 1 )

= x² + 2x( y + 1 ) + ( y + 1 )²

= ( x + y + 1 )²

= [ ( x + y ) + 1 ]²

= ( x + y )² + 2( x + y ) + 1

= x² + 2xy + y² + 2x + 2y + 1

\(a)\)

\(21\left(x+3\right)^3:\left(3x+9\right)^2\)

\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)

\(=7\left(x+3\right):3\)

Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)

\(b)\)

Thay vào ta được:

\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)

\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)

\(=1^4:\left(1^3.1\right)\)

\(=1:1\)

\(=1\)

\(c)\)

Thay vào ta được:

\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)

\(=-6.10.7\)

\(=-420\)

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

a) A=x^3 + 3x^2*5 + 3x*5^2 + 5^3

       =(x+5)^3

Thay x = -10 vào biểu thức A ta được:

A  = (-10+5)^3

    =(-5)^3

    =-75

Làm tương tự nhé

f) = x²( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )

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

h) = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

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

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

Trả lời:

f, x³ - 4x² - 9x + 36 = ( x³ - 4x² ) - ( 9x - 36 ) = x² ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x² - 9 ) = ( x - 4 )( x - 3 )( x + 3 )

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

h, x⁴ + x³ + x² - 1 = ( x⁴ + x³ ) + ( x² - 1 ) =  x³ ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 ) 

i, x² - y² - 4x - 4y = ( x² - y² ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )

j, x³ - y³ - 3x + 3y = ( x³ - y³ ) - ( 3x - 3y ) = ( x - y )( x² + xy + y² ) - 3 ( x - y ) = ( x - y )( x² + xy + y² - 3 ) 

B1 : a, M = x³-3xy(x-y)-y³-x²+2xy-y²

= ( x³-y³)-3xy(x-y) -(x²-2xy+y²)

= (x-y)(x²+xy+y²)-3xy(x-y)-(x-y)²

= (x-y) [(x²+xy+y²-3xy-(x-y)]

= (x-y)[(x²-2xy+y²)-(x-y)

= (x-y)[(x-y)²-(x-y)]

= (x-y)(x-y)(x-y-1)

= (x-y)²(x-y-1)

= 7²(7-1) = 49 . 6= 294

N = x²(x+1)-y²(y-1)+xy-3xy(x-y+1)-95

= x³+x²-(y³-y²)+xy-(3x²y-3xy²+3xy)-95

= x³+x²-y³+y²+xy-3x²y+3xy²-3xy-95

= (x³-y³)+(x²-2xy+y²)-(3x²y+y²)-(3x²y-3xy²)-95

=(x-y)(x²+xy+y²)+(x-y)²-3xy(x-y)-95

= (x-y)(x²+xy+y²+x-y-3xy)-95

= (x-y)[(x²-2xy+y²)+(x-y)]-95

= (x-y)[(x-y)²+(x-y)]-95

=(x-y)(x-y)(x-y+1)-95

= (x-y)²(x-y+1)-95

= 7²(7+1)-95=297