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15 tháng 7 2019

1)\(n^2\left(n-1\right)\left(n+1\right)-\left(n^2+2\right)\left(n^2-2\right)=n^2\left(n^2-1\right)-\left(n^4-4\right)=n^4-n^2-n^4+4\)

\(=-n^2+4\)

2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)

\(=y^4-81-y^4+16=-65\)

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

\(=x^2+6x+9-4y^2-x^2+4y^2=6x+9\)

4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

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

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

Học tốt nha bạn !

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)

21 tháng 7 2018

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

\(=x^3+2x^2-9x+2x^2+4x-18\)

\(=x^3+4x^2-5x-18\)

\(\left(x^{2y}-6\right)\left(x^2-5\right)\)

\(=x^{4y}-5x^{2y}-6x^2+30\)

\(\left(x+y\right)\left(xy-4+y\right)\)

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

câu còn lại tương tự  nha

19 tháng 7 2018

1, \(3x^2y^2-6x^2y^3+9x^2y^2\)

\(\Leftrightarrow12x^2y^2-6x^2y^2\)

\(\Leftrightarrow3x^2y^2\left(4+2y\right)\)

19 tháng 7 2018

5x^2y^3 - 25x^3y^4 + 10x^3y^3

\(\Leftrightarrow5x^2y^3\left(1-5xy+2x\right)\)

19 tháng 10 2023

a) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)

= x² + 3xy - 3x³ + 2y³ - xy + 3x³

= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³

= x² + 2xy + 2y³

Tại x = 5 và y = 4

M = 5² + 2.5.4 + 2.4³

= 25 + 40 + 2.64

= 65 + 128

= 193

b) N = x²(x + y) - y(x² - y²)

= x³ + x²y - x²y + y³

= x³ + (x²y - x²y) + y³

= x³ + y³

Tại x = -6 và y = 8

N = (-6)³ + 8³

= -216 + 512

= 296

c) P = x² + 1/2 x + 1/16

= (x + 1/2)²

Tại x = 3/4 ta có:

P = (3/4 + 1/2)² = (5/4)² = 25/16

a: \(=x^3+2x^2-9x+2x^2+4x-18\)

\(=x^3+4x^2-5x-18\)

b: \(=x^4y-5x^2y-6x^2+30\)

c: \(=x^2y-4x+xy+xy^2-4y+y^2\)

d: \(=x^3y^2-\dfrac{1}{2}x^2y^2-x^2+\dfrac{1}{2}x+\dfrac{3}{4}x-\dfrac{3}{8}\)

\(=x^3y^2-\dfrac{1}{2}x^2y^2-x^2+\dfrac{5}{4}x-\dfrac{3}{8}\)

Bài 4 :

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

b)\(x^2y^2+1-x^2-y^2=\left(x^2y^2-x^2\right)-\left(y^2-1\right)=x^2\left(y^2-1\right)-\left(y^2-1\right)=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)

c) \(x^2-y^2-4x+4y=\left(x^2-y^2\right)-\left(4x-4y\right)=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)=\left(x-y\right)\left(x+y-4\right)\)

d)

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

e) Trùng câu d

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

Bài 5:

a) \(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)

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

Vậy ...

b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\

\(\Leftrightarrow2x-3=6\)

\(\Leftrightarrow x=\frac{9}{2}\)

vậy........

c) \(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)

Vậy

d) \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy ........

9 tháng 9 2020

a)   \(=5x^2+40x+80+4\left(x^2-10x+25\right)-9\left(x+4\right)\left(x-4\right)\)

\(=5x^2+40x+80+4x^2-40x+100-9x^2+144\)

\(=9x^2-9x^2+40x-40x+324\)

\(=324\)

b)   \(=x^2+4xy+4y^2+4x^2-4xy+y^2-5x^2+5y^2-10y^2+90\)

\(=5x^2-5x^2+10y^2-10y^2+\left(4xy-4xy\right)+90\)

\(=90\)

c)

\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)

\(=\left(2a^2-2a^2\right)+\left(2b^2-2b^2\right)+2c^2+4ab-4ab+2\left(ac+bc-ac-bc\right)\)

\(=2c^2\)

9 tháng 9 2020

a) 5( x + 4 )2 + 4( x - 5 )2 - 9( 4 + x )( x - 4 )

= 5( x2 + 8x + 16 ) + 4( x2 - 10x + 25 ) - 9( x2 - 16 )

= 5x2 + 40x + 80 + 4x2 - 40x + 100 - 9x2 + 144

= ( 5x2 + 4x2 - 9x2 ) + ( 40x - 40x ) + ( 80 + 100 + 144 )

= 324

b) ( x + 2y )2 + ( 2x - y )2 - 5( x + y )( x - y ) - 10( y + 3 )( y - 3 )

= x2 + 4xy + 4y2 + 4x2 - 4xy + y2 - 5( x2 - y2 ) - 10( y2 - 9 )

= x2 + 4xy + 4y2 + 4x2 - 4xy + y2 - 5x2 + 5y2 - 10y2 + 90

= ( x2 + 4x2 - 5x2 ) + ( 4xy - 4xy ) + ( 4x2 + y2 + 5y2 - 10y2 ) + 90

= 90

c) ( a + b + c )2 + ( a + b - c )2 - 2( a + b )2

= [ ( a + b ) + c ]2 + [ ( a + b ) - c ]2 - 2( a + b )2

=  ( a + b )2 + 2( a + b )c + c2 + ( a + b )2 - 2( a + b )c + c2 - 2( a + b )2

= [ ( a + b )2 + ( a + b )2 - 2( a + b )2 ] + [ 2( a + b )c - 2( a + b )c ] + ( c2 + c2 )

= 2c2

9 tháng 7 2018

Sorry mình mới học lớp 6, ko giúp đc