d)123 x 456+456 x 321 - 256 x444

= ( 123 + 321) x456 - 256 x 444

= 444 x456 -256 x444

=200 x 444 

=88800

\(\text{c) (39 -19) : (-2) + (34 - 22).5}\\ =20:\left(-2\right)+12.5\\ =\left(-10\right)+60=50\)

\(\text{d) 123.456 + 456.321 – 256.444}\\ =456.\left(123-256+444\right)\\ =456.311=141816\)

c, (39 -19) : (-2) + (34 - 22).5

= 20: (-2) + 12.5 =(-10 )+ 60= 50

d, 123.456 + 456.321 – 256.444

=456.(123+321) - 256.444

= 456. 444 -256. 444

=444.(456 - 256)= 444.200 =88800

\(2000^{2001}⋮5\)

mà \(2001^{2002}⋮̸5\)

nên \(A⋮̸5\)

\(=\left[\dfrac{9-3}{162}\cdot\dfrac{81}{17}+\dfrac{35}{34}\right]:\left(\dfrac{3}{5}+\dfrac{7}{102}\right)\cdot\dfrac{102}{5}+2017\)

\(=\left[\dfrac{6}{2}\cdot\dfrac{1}{17}+\dfrac{35}{34}\right]:\dfrac{341}{510}\cdot\dfrac{102}{5}+2017\)

\(=\dfrac{41}{34}\cdot\dfrac{510}{341}\cdot\dfrac{102}{5}+2017=\dfrac{12546}{341}+2017=\dfrac{700343}{341}\)

\(\left(125^3.7^5-175^5:5\right):2016^{2017}\)

= \(\left[\left(5^3\right)^3.7^5-\left(5^2.7\right)^5:5\right]:2016^{2017}\)

= \(\left[5^{3.3}.7^5-5^{2.5}.7^5:5\right]:2016^{2017}\)

= \(\left[5^9.7^5-5^{10}.7^5:5\right]:2016^{2017}\)

= \(\left[5^9.7^5-5^9.7^5\right]:2016^{2017}\)

= \(0:2016^{2017}\)

= \(0\)

\(\dfrac{125^3\cdot7^5-175^5:5}{2016^{2017}}\)

\(=\dfrac{5^9\cdot7^5-5^{10}\cdot7^5:5}{2016^{2017}}\)

\(=\dfrac{5^9\cdot7^5-5^9\cdot7^5}{2016^{2017}}\)

=0

bang 0 day ban oi

neu dung cu  

Bai nay minh cung gap trong thi toan violympic ne ma khong co giai duoc buon ghe huuuuu

Bam may tinh thi ra la math erro

Ban oi sao khong co mu len gi het vay co the may ban kia doc se khong hieu dau

Ta có: 
(175⁵ : 5 - 125³ . 7⁵) : 2016²⁰¹⁷
= [(5² .7)⁵ : 5 - (5³)³ . 7⁵] : 2016²⁰¹⁷
= (5¹⁰ .7⁵ : 5 - 5⁹ . 7⁵) : 20162017
= [5⁹ . 7⁵ . (5 : 5 - 1)] : 2016²⁰¹⁷
= (5⁹ . 7⁵ . 0) : 2016²⁰¹⁷
= 0

c) Ta có: \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)

\(=\left(\dfrac{2\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-\sqrt{5}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)

\(=\left(2-\sqrt{5}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)

=1

d) Ta có: \(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)

=-2