1. 

$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$

2.

$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$

3. Không phù hợp để tính nhanh 

4. 

$=15^8-(15^8-1)=1$

5.

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$

$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$

$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$

6:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)

\(\frac{1-\frac{6}{23}+\frac{6}{31}-\frac{6}{154}}{\frac{5}{2}-\frac{15}{23}+\frac{15}{31}-\frac{15}{154}}\)

\(=\frac{\frac{6}{6}-\frac{6}{23}+\frac{6}{31}-\frac{6}{154}}{\frac{15}{6}-\frac{15}{23}+\frac{15}{31}-\frac{15}{154}}\)

\(=\frac{6\left(\frac{1}{6}-\frac{1}{23}+\frac{1}{31}-\frac{1}{154}\right)}{15\left(\frac{1}{6}-\frac{1}{23}+\frac{1}{31}-\frac{1}{154}\right)}\)

\(=\frac{6}{15}=\frac{2}{5}\)

mk sửa đề tí nhé

67.[(-154)+154].(-33)

=67 . 0.(-33)=0.(-33)=0

67.(-154)+154.(-33)

=(-67).154+154.(-33)

=154.\([\left(-67\right)+\left(-33\right)]\)

=154.(-100)

=-15400

= 3/30 nhé bạn

\(\frac{3}{20}\)

k mk nhé 

bạn có thể trả lời đầy đủ ko

`a, 126^2 - 152 . 126 + 5776.`

`= 126^2 - 2 . 76 . 126 + 76^2`

`= (126+76)^2 = 202^2 = 40804`

`b, 3^8 . 5^8 - (15^4-1)(15^4+1)`

`= 15^8 - 15^8 + 1`

`= 1`

 k thế nào z

\(\frac{3}{20}\)

bạn giải chi tiết ra giúp mik

a) nếu trong 1 biểu thức toàn cộng và trừ hay toàn nhân và chia trong đó có dấu ngoặc thì khi bỏ ngoặc , nếu trước ngoặc có dấu cộng hay nhân thì ta giữ nguyên dấu trong ngoặc, nếu trước ngoặc có dấy trừ thì ta phải đổi tất cả dấu trong ngoặc 

b) \(\left(2017-154+36\right)-\left(6-154+30\right)\)

\(=2017-154+36-6+154-30\)

\(=2017+\left(-154\right)+36+\left(-6\right)+154+\left(-30\right)\)

\(=2017+\text{ }\left[\left(-154\right)+154\right]+\left[36+\left(-6\right)\right]+\left(-30\right)\)

\(=2017+0+30+\left(-30\right)\)

\(=2017+\text{ }\left[30+\left(-30\right)\right]\)

\(=2017+0\)

\(=2017\)

(2017 - 154 + 36) - (6 -154 + 30).

= 2017 - 154 +36 - 6 + 154 - 30

= - 154 + 154 + 2017 - 30 + 36 - 6

= 0 + 1987 + 30

= 2017

\(\frac{1}{10}\)+\(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)

=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)

=\(\frac{1}{3}\)(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))

=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+\(\frac{1}{17}\)-\(\frac{1}{20}\))

=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{20}\))

=\(\frac{1}{3}\).\(\frac{9}{20}\)

=\(\frac{3}{20}\)

Ta có: S = 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340

=> S = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 +1/14.17 +1/17.20
Nhân 2 vế với 3 và áp dụng công thức tách 1 phân số thành hiệu 2 phân số: x/n.(n + x) = 1/n - 1/(n + x)
=> 3.S = 3.(1/2.5 + 1/5.8 + 1/8.11 +1/11.14 +1/14.17 +1/17.20)
=> 3.S = 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 +3/14.17 +3/17.20
=> 3.S = 1/2 - 1/ 5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20
=> 3.S = 1/2 - 1/20
=> 3.S = 9/20
=> S = 3/20

thanks nhiều nhak