\(\frac{4}{9}x^4-16x^2\)
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a. \(9x^2+30x+25=\left(3x+5\right)^2\)
b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)
d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)
e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)
f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)
a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)
b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)
a) (4x)2 , (9x2y)2 ,
b) (3ab4)3 , (\(-\frac{1}{5}\)x3y2)
\(x^2+5x+\frac{25}{4}\)
\(x+\left(\frac{5}{2}\right)^2+2.\frac{5}{2}.x\)
\(\left(x+\frac{5}{2}\right)^2\)
\(\frac{k}{me}\)
2:
-8x^6-12x^4y-6x^2y^2-y^3
=-(8x^6+12x^4y+6x^2y^2+y^3)
=-(2x^2+y)^3
3:
=(1/3)^2-(2x-y)^2
=(1/3-2x+y)(1/3+2x-y)
a,16x2-9
=42x2-32
=(4x-3)(4x+3) HĐT thứ 3
b,9a2-25b2
=32a2-52b2
=(3a-5b)(3a+5b) HĐT thứ 3
c,81-y4
=32.32-y2.y2
=(32-y2)
=(3-y)(3+y) HĐT thứ 3
d,(2x+y)2-1
=(2x+y-1)(2x+y-1) HĐT thứ 3
e,(x+y+z)2-(x-y-z)2
cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha
a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)
b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)
c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)
d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
\(\frac{4}{9}x^4-16x^2=\left(\frac{2}{3}x^2\right)^2-\left(4x\right)^2=\left(\frac{2}{3}x^2+4x\right).\left(\frac{2}{3}x^2-4x\right)\)