Question

[Nguyễn Thùy Linh]

Giải hộ em đi ạ :333

Answer 1

[Athanasia Karrywang]

1A

\(a,\left(2x+3\right)^2=4x^2+12x+9\)

\(b,\left(6-3n\right)^2=36-36n+9n^2\)

\(c,\left(y-4\right)\left(y+4\right)=y^2-16\)

\(d,\left(\frac{a}{2}-4\right)^2=\frac{a^2}{4}-4a+16\)

1B

\(a,\left(x-\frac{3}{4}\right)^2=x^2-\frac{3}{2}x+\frac{9}{16}\)

\(b,\left(3t+1\right)^2=9t^2+6t+1\)

\(c,\left(3a+\frac{1}{3}\right)\left(\frac{1}{3}-3a\right)=\frac{1}{9}-9a^2\)

\(d,\left(a^2-2\right)^2=a^4-4a^2+4\)

2A

\(a,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8a}{3}y+16y^2\)

\(b,\left(\frac{1}{x}-\frac{3}{y}\right)^2=\frac{1}{x^2}-\frac{6}{xy}+\frac{9}{y^2}\)

\(c,\left(\frac{x}{2}-\frac{yz}{6}\right)\left(\frac{x}{2}+\frac{yz}{6}\right)=\frac{x^2}{4}-\frac{y^2x^2}{36}\)

\(d,\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=x^4-\frac{4}{25}y^2\)

2B

\(a,\left(y-2xy\right)^2=y^2-4xy^2+4x^2y^2\)

\(b,16\left(-\frac{1}{4}x+\frac{3}{5}y\right)^2=16\left(\frac{9}{25}y^2-\frac{3}{10}xy+\frac{1}{16}x^2\right)=\frac{144}{25}y^2-\frac{48}{10}xy+x^2\)

\(c,\left(-\frac{1}{3}ab^2+c^3\right)\left(-\frac{1}{3}ab^2-c^3\right)=\frac{1}{9}a^2b^4-c^6\)

\(d,\left(a+\frac{2}{3}\right)^2\left(a-\frac{2}{3}\right)^2=\left(a^2-\frac{4}{9}\right)^2=a^4-\frac{8}{9}a+\frac{16}{81}\)