a.

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

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

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

\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}-2}.1=0\)

b.

\(=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\left(\sqrt{2}+1\right)}\)

\(=\sqrt{\sqrt{2}-1}-\left(\sqrt{2}-1\right)\left(\sqrt{\sqrt{2}+1}\right)\)

\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}.\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}=0\)

a)= 6/5 : 3/4 = 6/5 . 4/3 = 24/15 
b) 2 - ( 3/6 + 5/6 ) = 2 - 8/6 = 2 - 4/3 = 2/1 - 4/3 = 6/3 - 4/3 = 2/3

Giải:

a) S=5²/1.6+5²/6.11+5²/11.16+5²/16.21+5²/21.26

    S=5.(5.1/6+5/6.11+5/11.16+5/16.21+5/21.26)

    S=5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26)

    S=5.(1/1-1/26)

    S=5.25/26

    S=125/26

b) (1-1/2).(1-1/3).(1-1/4).(1-1/5).....(1-1/19).(1-1/20)

=1/2.2/3.3/4.4/5.....18/19.19/20

=1.2.3.4.....18.19/2.3.4.5.....19.20

=1/20

Chúc bạn học tốt!

Mình chui !

a: \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)

\(=x^5y-\dfrac{1}{5}x^3y^3-x^2y\)

b: \(\left(\dfrac{1}{2}x-5\right)\left(x^2-2x+3\right)\)

\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)

\(=\dfrac{1}{3}x^3-6x^2+\dfrac{23}{2}x-15\)

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a)

\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)      

b)

\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ =  - 30\end{array}\)

a: \(=\dfrac{4+15}{20}\cdot\dfrac{1}{2}=\dfrac{19}{20}\cdot\dfrac{1}{2}=\dfrac{19}{40}\)

b: \(=\dfrac{13}{20}-\dfrac{1}{4}+\dfrac{5}{20}=\dfrac{18}{20}-\dfrac{1}{4}\)

\(=\dfrac{9}{10}-\dfrac{1}{4}=\dfrac{18}{20}-\dfrac{5}{20}=\dfrac{13}{20}\)

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