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a: \(25x^2-\dfrac{10}{3}xy+\dfrac{1}{9}y^2=\left(5x-\dfrac{1}{3}y\right)^2\)
b: \(25x^2-15x+\dfrac{9}{4}=\left(5x-\dfrac{3}{2}\right)^2\)
c: \(\left(2x+\dfrac{1}{2}y\right)\left(4x^2-xy+\dfrac{1}{4}y^2\right)=8x^3+\dfrac{1}{8}y^3\)
d: \(\left(x^2-\dfrac{2}{3}\right)\left(x^4+\dfrac{2}{3}x^2+\dfrac{4}{9}\right)=x^6-\dfrac{8}{27}\)
B1:
\(=x^2+2x-5x-10+3\left(x^2-2^2\right)-\left(9x^2-2.3x.\frac{1}{2}+\frac{1}{4}\right)+5x^2\)
\(=-10-12-\frac{1}{4}=-22\frac{1}{4}\)
Bài 1.
( x - 5 )( x + 2 ) + 3( x - 2 )( x + 2 ) - ( 3x - 1/2 )2 + 5x2
= x2 - 3x - 10 + 3( x2 - 4 ) - ( 9x2 - 3x + 1/4 ) + 5x2
= 6x2 -- 3x - 10 + 3x2 - 12 - 9x2 + 3x - 1/4
= -89/4 không phụ thuộc vào biến
=> đpcm
Bài 2 < mình viết luôn nhé >
a) ( x + 2y2 )2 = x2 + 4xy2 + 4y4
b) ( a - 5/2b )2 = a2 - 5ab + 25/4b2
c) ( m + 1/2 )2 = m2 + m + 1/4
d) x2 - 16y4 = ( x + 4y2 )( x - 4y2 )
e) 25a2 - 1/4b2 = ( 5a + 1/2b )( 5a - 1/2b )
a) \(x^2+4x+4=\left(x+2\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
bài 1 : điền vào chỗ chấm để đk khẳng định đúng :
a) (.x..+2y...)2=x2+..4y.+4y2
b) (.a..-.3b..)2=a2-6ab+.9b2..
c) (.m..+.\(\frac{1}{2}\)..)2=.m2..+m+1/4
d) 25a2-..\(\frac{1}{4}b\).=(.5a..+1/2b)(..5a..-1/2b)
e)(.2x...+.1..)^2 = 4x^2 +.4x..+1
g)(2-x)(.4..+.2x..+.x2..)=8-x^3
h) 16a^2 - ..9. = (..4a.+3)(..4a.-3)
f)25 - ..30y.+9y^2=(..5.+...3y.)^2
A=(a+1)(a+2)(a^2+4)(a-1)(a^2+1)(a-2)
A =(a+1)(a-1)(a+2)(a-2)(a^2+4)(a^2+1)
A =(a^2-1)(a^2+1)(a^2-4)(a^2+4)
A =(a^4-1)(a^4-16)
A =\(a^{16}-16\cdot a^4-a^4+16\)
A =\(a^{16}-17\cdot a^4+16\)
B=(a+2b-3c-d)(a+2b+3c+d)
B=[(a+2b)^2 - (3c +d)^2]
B=[a^2+4ab+4b^2-(9c^2+6cd+d^2)]
B=a^3+4ab+4b^2 - 9c^2 - 6cd - d^2
C=(1-x-2x^3+3x^2)(1-x+2x^3-3x^2)
C=[(1-x)^2-(2x^3-3x^2)^2]
C=[(1-2x+x^2) - (4x^6-12x^5+9x^4)]
C=[1-2x-x^2-4x^6+12x^5-9x^4]
C=-4x^6+12x^5-9x^4-x^2-2x+1
D=(a^6-3a^3+9)(a^3+3)
D=a^9+27
a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
a) ( m + 1/2) ^2 = m2 + m + 1/4
b) x2 - 16y4 = (x + 4y2)(x-4y2)
c) 25a2 - 1/4b2 = (5a + 1/2b)(5a - 1/2b)
(?-?)^2=a^2-6ab+?