phan tich da thuc 26+y2-2x3y-64
phan tich da thuc thanh nhan tu:x^8+64
x8+64
=(x4)2+16x4+82-16x4
=(x4+8)2-(4x2)2
=(x4+8-4x2)(x4+8+4x2)
phan tich da thuc thanh nhan tu x^3 - 64
\(x^3-64=x^3-4^3\)
\(\Rightarrow\left(x-4\right)\left(x^2+4x+4^2\right)\)
Ta có:\(x^3-64\)
\(=x^3-4^3\)
Áp dụng hằng đẳng thức:\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\Rightarrow x^3-4^3=\left(x-4\right)\left(x^2+4x+4^2\right)\)
Phan tich da thuc thanh nhan tu : x2 - 4x -y2+4
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x-y^2+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
Phan tich da thuc thanh nhan tu : x2 - 4x -y2+4
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
phan tich da thuc thanh nhan tu (x-2)^3+64
( x-2 ) ^3 + 4^3
= hằng đẳng thức thứ 6 nha pạn
\(\left(x-2\right)^3+64\)
\(=\left(x-2\right)^3+4^3\)
\(=\left(x-2+4\right)\left[\left(x-2\right)^2-4\left(x-2\right)+4^2\right]\)
\(=\left(x+2\right)\left(x^2-4x+4-4x+8+16\right)\)
\(=\left(x+2\right)\left(x^2-8x+28\right)\)
Tham khảo nhé~
phan tich da thuc thanh nhan tu
x4 +y4 +64
Sửa đề: x^4+64
x^4+64
=x^4+16x^2+64-16x^2
=(x^2+8)^2-(4x)^2
=(x^2-4x+8)(x^2+4x+8)
phan tich da thuc thanh nhan tu
a, x^4.y^4+64
3.7: Su dung cac hang dang thuc de phan tich cac da thuc sau thanh nhan tu:
a) -y2 + 1/9
b) x4 - 256
c) 9 (x - 3)2 - 4 (x + 1)2
d) 25x2 - 1/81 x2y2
a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)
\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)
\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)
\(=\left(5x-7\right)\left(x-11\right)\)
\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)
phan tich cac da thuc sau thanh phan tu
6x^2+12xy+5x+10y
\(=6x\left(x+2y\right)+5\left(x+2y\right)=\left(6x+5\right)\left(x+2y\right)\)
\(=6x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(6x+5\right)\)