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a)\(9m^2+n^2-6mn\)
\(=9m^2-6mn+n^2\)
\(=\left(3m\right)^2-2.3m.n+n^2\)
\(=\left(3m-n\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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!!!
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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!
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a ) \(\left(a^6-3a^3+9\right)\left(a^3+3\right)=a^9+27\)
b ) Đặt \(a-y=t\) , ta có :
\(\left(t-x\right)^3-\left(t+x\right)^3\)
\(=\left(t-x-t-x\right)\left[\left(t-x\right)^2+\left(t-x\right)\left(t+x\right)+\left(t+x\right)^2\right]\)
\(=-2x\left[t^2-2tx+x^2+t^2-x^2+t^2+2tx+x^2\right]\)
\(=-2x\left[\left(t^2+t^2+t^2\right)+\left(x^2-x^2+x^2\right)+\left(2tx-2tx\right)\right]\)
\(=-2x\left(3t^2+x^2\right)\)
\(=-2x\left[3\left(a-y\right)^2+x^2\right]\)
\(=-2x\left(3a^2-6ay+3y^2+x^2\right)\)
c ) \(\left(4n^2-6mn+9m^2\right)\left(2n+3m\right)=8n^3+27m^3\)
d ) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)=125a^3-8b^3\)
a, ( a6 - 3a3 + 9 )(a3+ 3) = (a3)3 - 33 = a9 - 27
b, ( a-x-y)3 - (a+x-y)3 = (a-x-y-a+x-y)(a-x-y+a+x-y)
= (-2y)(2a-2y) = -2y.2(a-y)
c, (4n2- 6mn + 9m2)(2n + 3m) = (2n)3 + (3m)3
= 8n3 + 27m3
d, (25a2 + 10ab +4b2)( 5a - 2b ) = 125a3 - 8b3
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a) (a - 2b)2 = a2 - 2.a.2b + 4b2
= a2 - 4ab + 4b2
b) m2 - 4n2 = m2 - (2n)2 = (m - 2n)(m + 2n)
. Bài 1:
a; 9m^2 + n^2 - 6mn
= (3m)^2 - 2.3m.n + (n)^2
= ( 3m-n )^2
b; x^2-x+1/4
= x^2-2.(x).1/2+(1/2)^2
= (x-1/2)^2
![](https://rs.olm.vn/images/avt/0.png?1311)
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.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)
=\(\left(x-\frac{1}{2}\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
\(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2=\left(3m-n\right)^2\)
\(9m^2+n^2-6mn=\left(n-3m\right)^2=\left(3m-n\right)^2\)
đề bắt làm gì? => tùy theo yêu cầu cụ thể => đáp số.
ví dụ: đề là pt nhân tử => xong
đề là tìm nghiệm đa thức ....>= nghiệm duy nhất n =3m ......