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Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
a) \(4x^2+4xy+y^2=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
b) \(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2=\left(3m-n\right)^2\)
c) \(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2=\left(4a+5b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}=\left(x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\right)=\left(x-\dfrac{1}{2}\right)^2\)
Bài1:
\(\left(3+xy^2\right)^2=81+6xy^2+x^2y^4\)
Các câu sau tương tự
Bài2:
\(a,\left(4x^2+4xy+y^2\right)\)
=\(\left(2x+y\right)^2\)
b)\(9m^2+n^2-6mn=\left(3m-n\right)^2\)
c)\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
d)\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Bài3:
\(a,301^2=\left(300+1\right)^2=900+600+1=1501\)
b/\(499^2=\left(500-1\right)^2=2500-1000+1=1501\)
c/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
a) 4x2+4xy+y2
=(2x)2+2(2x)(y)+y2
=(2x+y)2
b)9m2+n2-6mn
=(3m)2-2(3m)n+n2
=(3m-n)2
c)16a2+25b2+40ab
=(4a)2+(5b)2+2(4a)(5b)
=(4a+5b)2
d)x2-x+\(\frac{1}{4}\)
=x2-2x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)
=\(\left(x-\frac{1}{2}\right)^2\)
a, \(4x^2+4xy+y^2=\left(4x\right)^2+2.2x.y+y^2\)
\(=\left(4x+y\right)^2\)
b, \(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2\)
\(=\left(3m-n\right)^2\)
c, \(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2\)
\(=\left(4a+5b\right)^2\)
d, \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
Chúc bạn học tốt!!!
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)
b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)
d)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
a) \(x^2+4x+4\)
\(=x^2+2\cdot2\cdot x+2^2\)
\(=\left(x+2\right)^2\)
b) \(4x^2-4x+1\)
\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)
\(=\left(2x-1\right)^2\)
c) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)
\(=\left[2\left(x+y\right)-1\right]^2\)
\(=\left(2x+2y-1\right)^2\)
Bài 2 :
\(4x^2+4xy+y^2=\left(2x+y\right)^2\)
\(9m^2+n^2-6mn=\left(3m-n\right)^2\)
\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Bài 1:
a, ( 3 + xy2)2 = 32 + 2. 3. xy2 + (xy2)2 = 9+ 6xy2 + x2y4.
b, (10- 2m2n) = 102 - 2.10.2m2n + (2m2n)2 = 100 - 40m2n + 4m4n
c, ( a - b2)(a+b2) = bạn xem lại đề câu này nhé!
Bài 2:
a, 4x2 + 4xy + y2 = (2x)2 + 2. 2x. y + y2 = ( 2x + y)2
b, 9m2 + n2 - 6mn = ( 3m)2 - 2. 3m. n + n2 = ( 3m - n)2
c,16a2 + 25b2 +40ab = (4a)2 + 2. 4a. 5b + (5b)2 = ( 4a + 5b) 2
d, x2 - x +1/4 = x2 - 2. 1/2. x +(1/2)2 = (x - 1/2)2