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: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)

a: \(=-4x^2+20x+2x-10=-4x^2+22x-10\)

b: =x^2-9

c: =x^3+27

d: \(=-2x^2-6x+x+3=-2x^2-5x+3\)

e: =8a^3+1

f: =(3-x)(x+1)(x+2)

=(3-x)(x^2+3x+2)

=3x^2+9x+6-x^3-3x^2-2x

=-x^3+7x+6

a.

\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)

\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)

b.

\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)

c.

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)

\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)

\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)

\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)

\(\left(2a+b-5\right)\left(2a-b+5\right)\)

\(=\left[2a+\left(b-5\right)\right]\left[2a-\left(b-5\right)\right]\)

\(=4a^2-\left(b-5\right)^2\)

\(=4a^2-\left(b^2-10b+25\right)\)

\(=4a^2-b^2+10b-25\)

\(\left(2a+b-5\right)\left(2a-b+5\right)\)

\(=\left(2a\right)^2-\left(b-5\right)^2\)

\(=4a^2-\left(b-5\right)^2\)