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\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)
\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)
a:Sửa đề: \(\dfrac{1}{4}a^2+2ab+4b^2\)
\(=\left(\dfrac{1}{2}a\right)^2+2\cdot\dfrac{1}{2}a\cdot2b+\left(2b\right)^2\)
\(=\left(\dfrac{1}{2}a+2b\right)^2\)
b: Sửa đề:\(y^4-\dfrac{1}{3}y^4+\dfrac{1}{36}\)
\(=y^8-2\cdot y^4\cdot\dfrac{1}{6}+\dfrac{1}{36}\)
\(=\left(y^4-\dfrac{1}{6}\right)^2\)
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) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)
\(=\left[2\left(x+y\right)-1\right]^2\)
\(=\left(2x+2y-1\right)^2\)
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)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)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
\(a,9a^2-6ab=1\) ( kiểm tra lại đề giúp mk)
\(b,25-10x+x^2\)
\(=5^2-2.5.x+x^2\)
\(=\left(5-x\right)^2=\left(x-5\right)^2\)
c, cx kiểm tra và viết rõ đề hộ mk ak
\(d,\left(x-y\right)^2-4\left(x-y\right)+4\)
\(=\left(x-y\right)^2-2.\left(x-y\right).2+2^2\)
\(=\left(x-y-2\right)^2\)
c/ \(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2\)
a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)
Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))
\(\frac{1}{4}a^2+2ab+4b^2\)
\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a+2b\right)^2\)
b. \(25+10x+x^2\)
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)
\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)
\(=\left(y^4-\frac{1}{3}\right)^2\)
c) (xy^2+1)^2
d) (1/3-y^4)^2
e) (1/2a-2b^2)^2
f) (5-x)^2
c) 2xy2 + x2 y4 + 1
= (xy2)2 + 2xy2 .1 + 12
=(xy2 +1)2
d) 1/9 - 2/3 y4 + y8
= (1/3)2 - 2 . 1/3 y4 + (y4)2
=(1/3 - y4)2
e) 1/4 a2 - 2ab2 + 4b4
= (1/2 a)2 - 2 1/2 a . 2b + (2b2)2
=(1/2 a - 2b2)2
f) 25 - 10x + x2
= x2 + 2 . 5x + 52
= (x+5)2