bài 1 rút gọn các biểu thức sau:
a) 5(x+4)^2+4(x-5)^2-9(4+x).(x-4)
b) (x+2y)^2+(2x-y)^2-5(x+y).(x-y)-10.(y+3).(y-3)
c) (a+b+c)^2+(a+b-c)^^2-2.(a+b)^2
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Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)
b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)
c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)
d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)
`a, (3x^2y)/(2xy^5)`
`= (3x)/(2y^4)`
`b, (3x^2-3x)/(x-1)`
`= (3x(x-1))/(x-1)`
`= 3x`
`c, (ab^2-a^2b)/(2a^2+a)`
`= (b(a-b))/((2a+1))`
`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
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
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
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) \(=5x^2+40x+80+4\left(x^2-10x+25\right)-9\left(x+4\right)\left(x-4\right)\)
\(=5x^2+40x+80+4x^2-40x+100-9x^2+144\)
\(=9x^2-9x^2+40x-40x+324\)
\(=324\)
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)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)
\(=\left(2a^2-2a^2\right)+\left(2b^2-2b^2\right)+2c^2+4ab-4ab+2\left(ac+bc-ac-bc\right)\)
\(=2c^2\)
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