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) \(x^2-x+\dfrac{1}{4}=\left(x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\right)=\left(x-\dfrac{1}{2}\right)^2\)

a,(2x+y)²

b,(3m-n)²

c,(4a+5b)²

d,(x-\(\dfrac{1}{2}\))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/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)

Bài 1:

a, \(\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)

b, \(\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)

c, \(\left(a-b\right)^2.\left(a+b\right)^2=\left[\left(a-b\right)\left(a+b\right)\right]^2\)

\(=\left(a^2-b^2\right)^2=a^4-2a^2b^2+b^4\)

Chúc bạn học tôt!!!

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!!!

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)

c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

Câu d thì biểu thức là \(\frac{x^2-1}{2x+\frac{1}{10}}\) hay là \(\frac{x^2-1}{\frac{2x+1}{10}}\) z bạn???

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) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

b) \(x^2+12xy+36y^2=\left(x+6y\right)^2\)

c) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

d) Không phải hằng đẳng thức \(\left(x^2-2x+4=\left(x-1\right)^2+3\right)\)

e) \(25x^2+4y^2-20xy=\left(5x-2y\right)^2\)