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

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

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

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

#Ayumu

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

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)

1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

=1

1

\(\left[\left(7^5:7^3+3^3\cdot20\right)^0\right]^{2016}\)

\(=1^{2016}\)

=1

`a)3/5+13/5-5/9-3/7`

`=16/5-5/9-3/7`

`=698/315`

3/5 + 13 /5 -5/9 - 3/7 = 698/315

bai EZ​ qua

mà bài dài quá sao làm hết dc

a, 5/6

b, 4/35

c, 0.6

5/6

4/35

0.6

=5/6

= 4/35

=0.6

1, 3/5 + -4/15 = \(\dfrac{9}{15}\)+ \(\dfrac{-4}{15}\)= \(\dfrac{5}{15}\)= \(\dfrac{1}{3}\)

2, -1/3 + 2/5 + 2/15= \(\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}\)=\(\dfrac{3}{15}=\dfrac{1}{5}\)

3, -3/5 + 7/21 + -4/5 + 7/5= \(\dfrac{1}{3}\)

\(1.\dfrac{3}{5}+\dfrac{-4}{15}=\dfrac{9}{15}+\dfrac{-4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

\(2.\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}=\dfrac{3}{15}=\dfrac{1}{5}\)

\(3.\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}=\left(\dfrac{-3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}=0+\dfrac{7}{21}=\dfrac{1}{3}\)

\(1,\dfrac{3}{5}+\dfrac{-4}{15}=\dfrac{9}{15}+\dfrac{-4}{15}\\ =\dfrac{9-4}{15}=\dfrac{1}{3}\\ 2,\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}\\ =\dfrac{3}{15}=\dfrac{1}{5}\)

^{Đã Làm !?}