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a) \(16a^2+25b^2+40ab=\left(4a\right)^2+\left(5b\right)^2+2\cdot4a\cdot5b=\left(4a+5b\right)^2\)
b)\(x^2-x+\frac{1}{4}=x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
Đây là rút gọn chứ ko phải tính
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) \(4x^2+4xy+y^2=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
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) \(x^2-x+\dfrac{1}{4}=\left(x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\right)=\left(x-\dfrac{1}{2}\right)^2\)
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) 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: \(9x^2-12xy+16y^2=\left(3x-4y\right)^2>=0\)(luôn đúng)
b: \(16a^2-30ab+25b^2=\left(4a-5b\right)^2>=0\)(luôn đúng)
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/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
16a2 + 25b2 + 40ab = (4a + 5b)2
\(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2=\left(4a+5b\right)^2\)
Chúc bn học tốt nha!