Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
Bài 1:
a: Ta có: \(M=\left(2a+b\right)^2-\left(b-2a\right)^2\)
\(=4a^2+4ab+b^2-b^2+4ab-4a^2\)
\(=8ab\)
b: Ta có: \(N=\left(3a+2\right)^2+2a\left(1-2b\right)+\left(2b-1\right)^2\)
\(=\left(3a+2+1-2b\right)^2\)
\(=\left(3a-2b+3\right)^2\)
\(=9a^2+4b^2+9-12ab+18a-12b\)
c: Ta có: \(A=\left(m-n\right)^2+4nm\)
\(=m^2-2mn+n^2+4mn\)
\(=m^2+2mn+n^2\)
\(=\left(m+n\right)^2\)
2:
a: \(\left(x+5\right)^2=x^2+10x+25\)
b: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
Bài 2:
c: \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d: \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e: \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f: \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
