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6 tháng 10 2020
5.
P = ( x - 1 )( x + 2 )( x + 3 )( x + 6 ) < sửa rồi nhé :v >
= [ ( x - 1 )( x + 6 ) ][ ( x + 2 )( x + 3 ) ]
= ( x2 + 5x - 6 )( x2 + 5x + 6 ) (1)
Đặt t = x2 + 5x
(1) = ( t - 6 )( t + 6 )
= t2 - 36 ≥ -36 ∀ t
Dấu "=" xảy ra khi t = 0
=> x2 + 5x = 0
=> x( x + 5 ) = 0
=> x = 0 hoặc x = -5
=> MinP = -36 <=> x = 0 hoặc x = -5
6 tháng 10 2020
6.
a) ( x2 + x )2 + 4( x2 + x ) = 12
Đặt t = x2 + x
pt <=> t2 + 4t = 12
<=> t2 + 4t - 12 = 0
<=> t2 - 2t + 6t - 12 = 0
<=> t( t - 2 ) + 6( t - 2 ) = 0
<=> ( t - 2 )( t + 6 ) = 0
<=> ( x2 + x - 2 )( x2 + x + 6 ) = 0
<=> x2 + x - 2 = 0 hoặc x2 + x + 6 = 0
+) x2 + x - 2 = 0
=> x2 - x + 2x - 2 = 0
=> x( x - 1 ) + 2( x - 1 ) = 0
=> ( x - 1 )( x + 2 ) = 0
=> x = 1 hoặc x = -2
+) x2 + x + 6 = ( x2 + x + 1/4 ) + 23/4 = ( x + 1/2 )2 + 23/4 ≥ 23/4 > 0 ∀ x
=> x ∈ { -2 ; 1 }
b) x2 - 12x + 36 = 81
<=> ( x - 6 )2 = ( ±9 )2
<=> x - 6 = 9 hoặc x - 6 = -9
<=> x = 15 hoặc x = -3
23 tháng 10 2019
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
B
9 tháng 9 2020
Bài làm :
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
X
14 tháng 7 2015
1,a4 + a3b + a + b
=a3(a+b)+(a+b)
=(a+b)(a3+1)
=(a+b)(a+1)(a2-a+1)
2,x3 + 2x2 - 2x - 1
=(x3-1)+(2x2-2x)
=(x-1)(x2+x+1)+2x(x-1)
=(x-1)(x2+x+1+2x)
=(x-1)(x2+3x+1)
3,4x(x-3y) + 12y(3y-x)
=-4x(3y-x)+12y(3y-x)
=(3y-x)(-4x+12y)
4,(x2-8) + 36
=x2-8-36+72
=(x2-62)+64
=(x-6)(x+6)+64
5,(x + 2) + (x + 3) + (x + 4) + (x + 5) - 24
=x + 2 + x + 3 + x + 4 + x + 5 -24
=4x-10
6,x2 - 2xy + y2 + 3x - 3y -10
=(x-y)2+3(x-y)-10
=(x-y)(x-y+3)-10
18 tháng 10 2020
A) 7X2 - 7Y2 - 14X + 14Y
= ( 7X2 - 7Y2 ) - ( 14X - 14Y )
= 7( X2 - Y2 ) - 14( X - Y )
= 7( X - Y )( X + Y ) - 14( X - Y )
= 7( X - Y )( X + Y - 14 )
B) X2 - Y2 + 14X + 49
= ( X2 + 14X + 49 ) - Y2
= ( X + 7 )2 - Y2
= ( X - Y + 7 )( X + Y + 7 )
C) X2 - Y2 - X + Y
= ( X2 - Y2 ) - ( X - Y )
= ( X - Y )( X + Y ) - ( X - Y )
= ( X - Y )( X + Y - 1 )
D) X2 + 12Y - Y2 - 36
= X2 - ( Y2 - 12Y + 36 )
= X2 - ( Y - 6 )2
= ( X - Y + 6 )( X + Y - 6 )
E) X3 + X2 - 9X - 9
= ( X3 + X2 ) - ( 9X + 9 )
= X2( X + 1 ) - 9( X + 1 )
= ( X + 1 )( X2 - 9 )
= ( X + 1 )( X - 3 )( X + 3 )
HT
13 tháng 10 2018
\(A=\dfrac{2x^3-18x}{x^4-81}\\ A=\dfrac{2x\left(x^2-9\right)}{\left(x^2+9\right)\left(x^2-9\right)}\\ A=\dfrac{2x}{x^2+9}\)
\(B=\dfrac{x^2-x-20}{x^2-25}\\ B=\dfrac{x\left(x-5\right)+4\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{\left(x-5\right)\left(x+4\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{x+4}{x+5}\)
\(C=\dfrac{8xy-6x^2}{12y^2-9xy}\\ C=\dfrac{2x\left(4y-3x\right)}{3y\left(4y-3x\right)}\\ C=\dfrac{2x}{3y}\)
\(E=\dfrac{x^2+5x+6}{x^2-4}\\ E=\dfrac{x\left(x+2\right)+3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{x+3}{x-2}\)
HP
3
26 tháng 10 2018
a) \(2x^2-50\)
\(=2\left(x^2-25\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b) \(x^2z+4xyz+4y^2z\)
\(=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c) \(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-2\cdot x\cdot6+6^2\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d) Đặt \(D=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(D=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(D=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x+1=a\)
\(D=\left(a-1\right)\left(a+1\right)+1\)
\(D=a^2-1^2+1\)
\(D=a^2\)
Thay \(x^2+3x+1=a\)vào D ta có :
\(D=\left(x^2+3x+1\right)^2\)
26 tháng 10 2018
a: \(2x^2-50=2\left(x^2-5^2\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b:\(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c:\(x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d:\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x+1\)Ta có:
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1=t^2\)
\(=\left(x^2+3x+1\right)^2\)
8 tháng 4 2018
a) \(x^4+2x^3-12x^2-13x+42=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)
ND
8 tháng 4 2018
\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
Ta có:
\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy...........