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B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
F=x2+2xy+y2-x-y-12
= (x + y)^2 - (x + y) - 12
= (x + y)(x + y - 1) - 12
đặt x + y = t
F = t(t - 1) - 12
= t^2 - t - 12
= (t - 4)(t + 3)
G=(x2-3x-1)2-12(x2-3x-1)+27
đăth x^2 - 3x - 1 = t
G = t^2 - 12t + 27
= (t - 3)(t - 9)
có t = x^2 - 3x - 1
thay vào
Câu F ( kiểm tra lại đề )
Câu G . Đặt x^2 -3x-1=t
t^2 -12t+27 ( thực hiện pp tách)
Đặt x2 + 4x + 8 = A. Ta sẽ được:
A2 + 3xA + 2x2
= A2 - xA - 2xA + 2x2
= A(A-x) - 2x(A-x)
= (A-x)(A-2x)
= (x2+3x+8)(x2+2x+8)
a) 4xn+2 + 8xn = 4xn( x2 + 2 )
b) ( 4x - 8 )( x2 + 6 ) - ( x - 2 )( x + 7 ) - 10 + 5x
= 4( x - 2 )( x2 + 6 ) - ( x - 2 )( x + 7 ) + 5( x - 2 )
= ( x - 2 )[ 4( x2 + 6 ) - ( x + 7 ) + 5 ]
= ( x - 2 )( 4x2 + 24 - x - 7 + 5 )
= ( x - 2 )( 4x2 - x + 22)
a) Đề đúng: \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=y\)
BT = \(y^2+4y-12\)
\(=\left(y+2\right)^2-4^2\)
\(=\left(y-2\right)\left(y+6\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+3\right)\)
b) Đặt \(x^2+x+1=y\)
=> BT = \(y\left(y+1\right)-12\)
\(=y^2+y-12\)
\(=\left(y-3\right)\left(y+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
d) x^3 + 2x^2 + 3x + 1
Bó tay
e) x^2 - 2x - 4y^2 - 4y
= x^2 - 2x + 1 - 4y^2 - 4y - 1
= ( x + 1 )^2 - ( 4y^2 + 4y + 1 )
= ( x + 1 )^2 - ( 2y + 1 )^2
= ( x+ 1 - 2y - 1 )( x + 1 + 2y + 1 )
= ( x - 2y )(x + 2y +2 )
4x2 là gì
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-2\right)\left(x+2y\right)\)
hk tốt
^^
Đặt x^2 + 2x = y thay vào ta có:
y(y+4) + 3 = y^2 + 4y +3 = y^2 + y + 3y + 3 = y(y+1) + 3(y + 1) = ( y + 3)( y+ 1)
Thay y = x^2 + 2x ta có
( x^2 + 2x + 3)(x^2 + 2x+ 1) = ( x^2 + 2x + 3) (x+ 1)^2
Đúng cho mình nha
\(\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
Đặt \(x^2+2x+2=t\)
\(\Rightarrow\left(t-2\right)\left(t+2\right)+3=t^2-4+3=t^2-1=\left(t-1\right)\left(t+1\right)\)
\(=\left(x^2+2x+2-1\right)\left(x^2+2x+2+1\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(=\left(x+1\right)^2.\left(x^2+2x+3\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)