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

\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)

a: \(\left(2x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(x-3\right)^2\)

\(=4x^2-4x+1-x^2+6x-9-3\left(x^2+x-2\right)\)

\(=3x^2+2x-8-3x^2-3x+6\)

=-x+2

b: \(\left(x-2\right)\left(2x-1\right)-3\left(x+1\right)^2-4x\left(x+2\right)\)

\(=2x^2-x-4x+2-3x^2-6x-3-4x^2-8x\)

\(=-5x^2-19x-1\)

a) (2x+3)²-2(2x+3)(2x+5)+(2x+5)²

=4x²+12x+9-(4x+6)(2x+5)+4x²+20x+25

=4x²+12x+9-(8x²+12x+20x+30)+4x²+20x+25

=4x²+12x+9-8x²-12x-20x-30+4x²+20x+25

=4

b) (x²+x+1)(x²-x+1)(x²-1)

=((x²+1)²-x²)(x²-1)

=(x⁴+x²+1)(x²-1)

=x⁶+x⁴+x²-x⁴-x²-1

=x⁶-1

c)(a+b-c)²+(a-b+c)²-2(b-c)²

=a²+b²+c²+2ab-2ac-2bc+a²+b²+c²-2ab+2ac-2bc-2(b²-2bc+c²)

=2a²+2b²+2c²-4bc-2b²+4bc-2c²

=2a²

d) (a+b+c)²+(a-b-c)²+(b-c-a)²+(c-a-b)²

= a²+b²+c²+2ab+2ac+2bc+a²+b²+c²-2ab-2ac+2bc+a²+b²+c²+2bc-2ab+2ac+a²+b²+c²-2ac-2bc+2ab

=4a²+4b²+4c²+4ab+4bc

a) Ta có: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)

\(=\left(2x+3-2x-5\right)^2\)

=4

b) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)

\(=\left(x^3-1\right)\left(x^3+1\right)\)

\(=x^6-1\)

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????????

\(\left(\dfrac{6-2}{1-3}-\dfrac{5}{5}\right):\dfrac{1}{5-2}\)

\(=\left(\dfrac{4}{-2}-1\right):\dfrac{1}{3}\)

\(=\left(-2-1\right):\dfrac{1}{3}\)

\(=-3.3\)

\(=-9\)

\(=\left(\dfrac{4}{-2}-1\right):\dfrac{1}{3}\)

=(-3)*3

=-9

đkxđ:\(x\ne5,x\ne-5\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5x+25}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)

thay x=1 vào bt A, ta được:

\(-\dfrac{4}{1-5}=1\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)