10  .tick cho minh nha

kết quả bang 50

ê! đây là môn toán mà!!

ai thấy đúng cho mk 1 tick nha!

\(\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)

đối với lớp 6 thì trình bày thế này dễ hiểu hơn

\(\frac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2^3.5.5^2.7.7^3}{2.5.7^2.2.5.7^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5}{1}=10\)

câu trả lời của mình là 10!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 

=10 nha k minh nha

Mình kb

Xin lỗi bạn nghe

=1,7.2,3+1,7.(-3,7)-1,7.3-1,7

=1,7.[2,3+(-3,7)-3-1]

=1,7.(-5,4)

= -9,18

a, \(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}=\frac{2\cdot2^2\cdot5^2.5.7.7^2}{2\cdot5\cdot7^2}=\frac{1\cdot2^2\cdot5^2\cdot7^2}{1\cdot1\cdot1}=4900\)

b, \(1,7\cdot2,3+1,7\cdot\left(-3,7\right)-1,7.3-1,7:0,1=1,7\cdot\left(-3,7+2,3-3\right)-1,7:0,1=1,7.\left(-4,4\right)-1,7:0,1=\left(-2,7\right)-17=-19,7\)

sai phần b rồi

\(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\frac{2^3\left(5\cdot5^2\right)\left(7\cdot7^3\right)}{2^2\cdot5^2\cdot7^4}\)

\(=\frac{2^2\cdot2\cdot5\cdot5^2\cdot7^4}{2^2\cdot5^2\cdot7^4}\)

Triệt tiêu ta còn \(2\cdot5=10\)

\(\frac{\left(2^3.5.7\right).\left(5^2.7^2\right)}{\left(2.5.7^2\right)^2}\)

\(=\frac{2^3.\left(5.7\right).\left(5^2.7^3\right)}{2^2.5^2.7^4}\)

\(=\frac{2^2.2.5.5^2.7.7^3}{2^2.5^2.7^4}\)

\(=\frac{2^2.2.5.5^2.7^4}{2^2.5^2.7^4}\)

\(=2.5\)

\(=10\)

A = 27 + 46 + 79 + 34 + 53 = 239

B = - 377 - (98 - 277)

= - 377 - 98 + 277

= -198

C = - 1,7× 2,3 + 1,7.× (- 3,7) - 1,7×3 - 0,17:0,1

= 1,7. (-2,3 – 3,7 – 3 – 1)

= 1,7 . (-10)

= -17.

D = $2\frac{3}{4}.\left(-0,4\right)-1\frac{3}{5}.2,75+\left(-1,2\right):\frac{4}{11}$

$=\frac{11}{4}.\frac{-4}{10}-\frac{8}{5}.\frac{11}{4}+\frac{-6}{5}.\frac{11}{4}$

$=\frac{11}{4}.\left(\frac{-4}{10}-\frac{8}{5}+\frac{-6}{5}\right)$

$=\frac{11}{4}.\frac{-2-8-6}{5}$

$=\frac{11}{4}.\frac{-16}{5}$

$=\frac{-44}{5}$

a=239

b=-198

c=-17

d=-44/5

e=10

\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{2.5.\left(7^2\right)}=\dfrac{2^3.\left(5.5^2\right).\left(7.7^3\right)}{2.5.7^2}\)

\(=\dfrac{2^3.5^3.7^4}{2.5.7^2}=\left(2^3:2\right).\left(5^3:5\right).\left(7^4:7^2\right)=2^2.5^2.7^2\)

=\(\left(2.5.7\right)^2=4900\)

(2³ . 5 . 7) . (5² . 7³) / 2 . 5 (7²)

= (8 . 5 . 7) . (25 . 343) / 2 . 5 . 49

= 280 . 8575 / 490

= 2401000 / 490

= 4900

=(8.5.7).(25.343)/10.49

=280.8575/490

=2401000/490

=490

Bài 1: 

a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)

\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)

\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)

\(=-10\cdot1.7=-17\)

b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)

\(=-\dfrac{539}{60}\)

c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)

\(=10\)