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a) \(x^2-10\cdot2\cdot x+10^2=\left(x-10\right)^2\)
b) \(x^2+2\cdot5\cdot x+5^2=\left(x+5\right)^2\)
c) \(x^2-2\cdot6\cdot xy+\left(6y\right)^2=\left(x-6y\right)^2\)
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) \(x^2+10x+26+y^2+2y\)
= \(x^2+10x+25+y^2+2y+1\)
= \(\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(x^2-2xy+2y^2+2y+1\)
= \(x^2-2xy+y^2+y^2+2y+1\)
= \(\left(x-y\right)^2+\left(y+1\right)^2\)
c) \(z^2-6z+5-t^2-4t\)
= \(z^2-6z+9-\left(t^2+4t+4\right)\)
= \(\left(z-3\right)^2-\left(t+2\right)^2\)
d) \(4x^2-12x-y^2+2y+1\)
Hình như câu này sai đề -_-
a, \(x^2+10x+26+y^2+2y\)
\(=\left(x^2+2.x.5+5^2\right)+\left(1^2+2.1.y+y^2\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b, \(x^2-2xy+2y^2+2y+1\)
\(=x^2-2xy+y^2+y^2+2y+1\)
\(=\left(x^2-2.x.y+y^2\right)+\left(y^2+2.y.1+1^2\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c,\(z^2 -6z+5-t^2-4t\)
\(=-\left(t^2+4t-z^2+6z-5\right)\)
\(=-\left(t^2+2.t.2+2^2-z^2+2.z.3-3^2\right)\)
\(=-\left(\left(t^2+2.t.2+2^2\right)-\left(z^2-2.z.3+3^2\right)\right)\)
\(=-\left(\left(t+2\right)^2-\left(z-3\right)^2\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
d, Không biết làm hihi :)
a. Để biểu thức là bình phương 1 hiệu thì
\(9x^2-30x+A=\left(3x\right)^2-2.3x.5+5^2=\left(3x-5\right)^2\)\(\Rightarrow A=25\)
b. Tương tự\(A-52xy^2+169y^4=\left(13y^2\right)^2-2.13.y^2.2x+4x^2=\left(13y^2-2x\right)^2\)
\(\Rightarrow A=4x^2\)
\(a,16x^2+24xy+.....\)
\(=\left(4x\right)^2+2.4x.3y+\left(3y\right)^2\)
\(=\left(4x+3y\right)^2\)
Vậy \(....=9y^2\)
\(b,25x^2+....+81\)
\(=\left(5x\right)^2+....+9^2\)
\(=\left(5x\right)^2+2.5x.9+9^2\)
\(=\left(5x+9\right)^2\)
Vậy \(....=90x\)
\(c,....-42xy+49y^2\)
\(=49y^2-42xy+....\)
\(=\left(7y\right)^2-2.7y.3x+\left(3x\right)^2\)
\(=\left(7y-3x\right)^2\)
\(=\left(3x-7y\right)^2\)
Vậy \(....=9x^2\)
\(2xy^2+x^2y^4+1\\ =\left(xy^2\right)^2+2xy^2.1+1^2\\ =\left(xy^2+1\right)^2\)
Ta có :
\(2xy^2+x^2y^4+1=\left(xy^2\right)^2+2.xy^2.1+1^2\)
\(=\left(xy^2+1\right)^2\)
1. 2xy2 +x2y4+1 = (xy2+1)2
2. a)3x2+3x-10x-10=3x(x+1)-10(x+1)=(x+1)(3x-10)
b)2x2-5x-7=2x2+2x-7x-7=2x(x+1)-7(x+1)=(x+1)(2x-7)
Mong có thể giúp được bạn
\(\frac{2a+1}{a^2\left(a+1\right)^2}=\frac{1}{a^2}-\frac{1}{\left(a+1\right)^2}\)