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\(81x^2-y^2=\left(9x\right)^2-y^2\)\(=\left(9x-y\right)\left(9x+y\right)\)
đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)
\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)
\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)
Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)
Ủng hộ = 1 cái t i c k nha cảm ơn
x^8+x^4+1=x^8-x^2+x^4-x+x^2+x+1=x^2(x^6-1)+x(x^3-1)+x^2+x+1=x^2(x^3-1)(x^3+1)+x(x^3-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)[x^2(x^3+1)(x-1)+x(x-1)+1)]
\(\left(6x-1\right)^2-\left(3x+2\right)\)
\(=36x^2-12x+1-3x-2\)
\(=36x^2-15x-1\)
bn ktra lại đề nhé
hk tốt
36(x-y)2-25(2x-y)2
= 36(x-y)2 - 100(x-y)2
=(36-100)(x-y)2
= -64(x-y)2
a) \(9\left(2x-3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x-3\right)-2\left(x+1\right)\right]\left[3\left(2x-3\right)+2\left(x+1\right)\right]\)
\(=\left(6x-9-2x-2\right)\left(6x-9+2x+2\right)\)
\(=\left(4x-11\right)\left(8x-7\right)\)
b) \(\left(x^2+4y^2-20\right)-16\left(xy-4\right)^2\)
\(=\left[\left(x^2-4xy+4y^2\right)-4\right]\left[\left(x^2+4xy+4y^2\right)-36\right]\)
\(=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]\)
\(=\left(x-2y-2\right)\left(x-2y+2\right)\left(x+2y-6\right)\left(x+2y+6\right)\)
a. 9 ( 2x - 3 )2 - 4 ( x + 1 )2
= [ 3 ( 2x - 3 ) ]2 - [ 2 ( x + 1 ) ]2
= [ 3 ( 2x - 3 ) - 2 ( x + 1 ) ] [ 3 ( 2x - 3 ) + 2 ( x + 1 ) ]
= ( 6x - 9 - 2x - 2 ) ( 6x - 9 + 2x + 2 )
= ( 4x - 11 ) ( 8x - 7 )
b. ( x2 + 4y2 - 20 )2 - 16 ( xy - 4 )2
= ( x2 + 4y2 - 20 )2 - [ 4 ( xy - 4 ) ]2
= [ x2 + 4y2 - 20 - 4 ( xy - 4 ) ] [ x2 + 4y2 - 20 + 4 ( xy - 4 ) ]
= ( x2 + 4y2 - 20 - 4xy + 16 ) ( x2 + 4y2 - 20 + 4xy - 16 )
= ( x2 + 4y2 - 4xy - 4 ) ( x2 + 4y2 + 4xy - 36 )
= [ ( x - 2y )2 - 22 ] [ ( x + 2y )2 - 62 ]
= ( x - 2y - 2 ) ( x - 2y + 2 ) ( x + 2y - 6 ) ( x + 2y + 6 )
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
\(=\left(4-a+b\right)\left(4+a-b\right)\)
\(=\left(4-a-b\right)\left(4+a-b\right)\), đằng trước là dấu trừ thì khi bỏ ngoặc phải đổi dấu chứ nhỉ :0