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b:=y^2+2y+1+9x^2-12x+4
=(y+1)^2+(3x-2)^2
a:
SỬa đề: 5y^2
=y^2-10y+25+9x^2+4y^2-12xy
=(y-5)^2+(3x-2y)^2
\(34+24\sqrt{2}=18+2\sqrt{288}+16=\left(\sqrt{18}\right)^2+2\sqrt{18}\cdot\sqrt{16}+\left(\sqrt{16}\right)^2=\left(\sqrt{18}+\sqrt{16}\right)^2\)
b)\(27-10\sqrt{2}=5^2-2.5\sqrt{2}+2=\left(5-\sqrt{2}\right)^2\)
c)\(18-8\sqrt{2}=4^2-2.4\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)
d)\(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
e)\(6\sqrt{5}+14=9+2.3\sqrt{5}+5=\left(3+\sqrt{5}\right)^2\)
f)\(20\sqrt{5}+45=5^2+2.5.2\sqrt{5}+20=\left(5+2\sqrt{5}\right)^2\)
g)\(7-2\sqrt{6}=6-2\sqrt{6}+1=\left(\sqrt{6}-1\right)^2\)
\(4x^2-\frac{1}{9}\left(y+1\right)^2=\left(2x\right)^2-\left(\frac{1}{3}\left(y+1\right)\right)^2\)
\(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)
\(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)
\(b,=\left(x^4y^8\right)^2+2\cdot2x^4y^8+2^2=\left(x^4y^8+2\right)^2\)
\(25a^2-20ab+4b^2\)
= \(\left(5a\right)^2\) \(-2.5a.2b\) \(+\left(2b\right)^2\)
= \(\left(5a-2b\right)^2\)
\(=\left(5a\right)^2-2\cdot5\cdot2\cdot a\cdot b+\left(2b\right)^2=\left(5a-2b\right)^2\)
A=9x^2−6x+1
=(3x)^2−2.3x.1+1^2
=(3x−1)^2
B=(2x+3y)^2+(2x+3y)+1(2x+3y)2+(2x+3y)+1
=[(2x+3y)^2+2.(2x+3y).1/2+(1/2)^2]+3/4
=(2x+3y+1/2)^2+3/4
=(2x+3y+1)(2x+3y)+1
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