Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

=> A=24497550

\(c,\dfrac{5}{18}\times\dfrac{2}{15}:\dfrac{1}{5}\\ =\dfrac{1}{9}\times\dfrac{1}{3}\times5\\ =\dfrac{5}{27}\\ d,\dfrac{4}{13}+\dfrac{2}{11}:\dfrac{4}{33}\\ =\dfrac{4}{13}+\dfrac{2}{11}\times\dfrac{33}{4}\\ =\dfrac{4}{13}+\dfrac{3}{2}=\dfrac{8}{26}+\dfrac{39}{26}\\ =\dfrac{47}{26}\)

\(c)\)\(\left(-\dfrac{1}{2}\right)^3:1\dfrac{3}{8}-25\%\cdot\left(-6\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{8}\cdot\dfrac{8}{11}-\dfrac{1}{4}\cdot\left(-\dfrac{68}{11}\right)\)

\(=-\dfrac{1}{11}+\dfrac{17}{11}\)

\(=\dfrac{16}{11}\)

#Ayumu

a) 1/5 - (1/2 + 3/4 ) : 5/2

= 1/5 - ( 1/4 + 3/4 ) : 5/2

=1/5 - 1 : 5/2

= 1/5 - 1 . 2/5

= 1/5 - 2/5

= -1/5

b) 1,5 . (1/3 - 2/3)

=3/2 . ( -1/3)

=-1/2

c) 9/10 . 23/11 - 1/11 . 9/10 + 9/10

= 9/10 . ( 23/11 - 1/11 ) + 9/10

= 9/10 . 1 + 9/10

= 9/10 + 9/10

= 18/10 = 9/5

Bài 1:
a. ĐKXĐ: $3x\geq 0$

$\Leftrightarrow x\geq 0$

b. ĐKXĐ: $\frac{x-1}{x+3}\geq 0$

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x-1\geq 0\\ x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} x-1\leq 0\\ x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 1\\ x< -3\end{matrix}\right.\)

Bài 2:

\(C=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{2+2\sqrt{2.3}+3}-\sqrt{2-2\sqrt{2.3}+3}\)

\(=\sqrt{(\sqrt{2}+\sqrt{3})^2}-\sqrt{(\sqrt{2}-\sqrt{3})^2}\)

\(=|\sqrt{2}+\sqrt{3}|-|\sqrt{2}-\sqrt{3}|=(\sqrt{2}+\sqrt{3})-(\sqrt{3}-\sqrt{2})\)

\(=2\sqrt{2}\)

Bài 3:
a. 

\(A=\frac{2}{\sqrt{x}+2}:\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]\)

\(=\frac{2}{\sqrt{x}+2}:\frac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{2}{\sqrt{x}+2}.\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}}=\frac{2(\sqrt{x}-2)}{\sqrt{x}}\)

b. Khi $x=\frac{1}{4}$ thì $\sqrt{x}=\frac{1}{2}$.

Khi đó $A=\frac{2(\frac{1}{2}-2)}{\frac{1}{2}}=-6$

c.

$A=\frac{2(\sqrt{x}-2)}{\sqrt{x}}=2-\frac{4}{\sqrt{x}}$

$< 2$ do $\frac{4}{\sqrt{x}}>0$

Ta có đpcm

d. Với $x$ nguyên, để $A$ nguyên thì $\sqrt{x}$ là ước của $4$

$\Leftrightarrow \sqrt{x}\in\left\{1;2;4\right\}$

$\Rightarrow x\in\left\{1;4;16\right\}$ (đều tm)

a)=-7/21+8/24

=-1/3+1/3

=0

b)=-3/5.(2/7+5/7)+23/5

=-3/5.7/7+23/5

=-3/5.1+23/5

=-3/5+23/5

=20/5=4

c)=75/100-11/2+5/10:5/12+1/4

=3/4-11/2+1/2:5/12+1/4

=3/4+-11/2+1/2.12/5+1/4

=3/4+-22/4+6/5+1/4

=-19/4+6/5+1/4

=(-19/4+1/4)+6/5

=-18/4+6/5

=-9/2+6/5

=-45/10+12/10

=-23/10

\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)

- \(\dfrac{3}{4}\).(-\(\dfrac{55}{9}\)).\(\dfrac{8}{11}\)

= \(\dfrac{3.5.11.4.2}{4.3.3.11}\)

= \(\dfrac{10}{3}\)

   1\(\dfrac{4}{23}\) + (\(\dfrac{5}{21}\) - \(\dfrac{4}{23}\)) + \(\dfrac{16}{21}\) - \(\dfrac{1}{2}\)

= 1 + \(\dfrac{4}{23}\) + \(\dfrac{5}{21}\) - \(\dfrac{4}{23}\) + \(\dfrac{16}{21}\) - \(\dfrac{1}{2}\)

= 1 + (\(\dfrac{4}{23}\) - \(\dfrac{4}{23}\)) + (\(\dfrac{5}{21}\) + \(\dfrac{16}{21}\)) - \(\dfrac{1}{2}\)

= 1 + 0 + 1 - \(\dfrac{1}{2}\)

= 2  - \(\dfrac{1}{2}\)

= \(\dfrac{3}{2}\)

- \(\dfrac{3}{8}\).\(\dfrac{6}{13}\) + \(\dfrac{7}{13}\). (-\(\dfrac{3}{8}\)) + 1\(\dfrac{3}{8}\)

= -\(\dfrac{3}{8}\).(\(\dfrac{6}{13}\) + \(\dfrac{7}{13}\)) + 1 + \(\dfrac{3}{8}\)

= - \(\dfrac{3}{8}\) + 1 + \(\dfrac{3}{8}\)

= (-\(\dfrac{3}{8}\) + \(\dfrac{3}{8}\)) + 1

= 0 + 1

= 1