(12.4⁴-2.4⁵):4⁵

=4⁴.(12-2.4):4⁵

=4⁴.4:4⁵

=4⁵:4⁵

=4⁰

=1

Chúc bạn học giỏi nha!!!

K cho mik với nhé Vũ Minh Hiếu

\(\frac{\left(3.4.4^4-2.4^5\right)}{4^5}=\frac{\left(3.4^5-2.4^5\right)}{4^5}=\frac{4^5.\left(3-2\right)}{4^5}=1.\)

( 12.4⁴- 2.4⁵ ) :4⁵
= ( 3. 4⁵- 2.4⁵):4⁵
=(4⁵.(3-2) ) :4⁵
=4⁵:4⁵
=1

a) (32+96):8=128:8=16
b) (3072-2048):1024=1024-1024=0

a: =5+48=53

a; -\(\dfrac{4}{9}.\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\)

= - \(\dfrac{4}{9}\).(\(\dfrac{7}{15}+\dfrac{8}{15}\))

= - \(\dfrac{4}{9}\).1

= - \(\dfrac{4}{9}\)

b; - \(\dfrac{5}{4}.\dfrac{16}{25}\) + \(\dfrac{-5}{4}\).\(\dfrac{9}{25}\)

= - \(\dfrac{5}{4}\).(\(\dfrac{16}{25}\) + \(\dfrac{9}{25}\))

= - \(\dfrac{5}{4}\). 1

= - \(\dfrac{5}{4}\)

c; \(-\dfrac{5}{12}.\dfrac{4}{19}\) + \(\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)

= - \(\dfrac{5}{12}\).\(\dfrac{4}{19}\) - \(\dfrac{7}{12}\).\(\dfrac{4}{19}\) - \(\dfrac{4}{19}\).\(\dfrac{10}{3}\)

= - \(\dfrac{4}{19}\).(\(\dfrac{5}{12}\) + \(\dfrac{7}{12}\) + \(\dfrac{10}{3}\))

= - \(\dfrac{4}{19}\).\(\dfrac{13}{3}\)

= - \(\dfrac{52}{57}\)

Đề thiếu dữ kiện. Bạn xem lại đề.

Lời giải:

$S=1+2.4+3.4^2+....+12.4^{11}$

$4S=1.4+2.4^2+3.4^3+....+12.4^{12}$

$\Rightarrow 3S=4S-S=12.4^{12}-(1+4+4^2+4^3+...+4^{11})$

$12S=12.4^{13}-(4+4^2+4^3+...+4^{12})$

$\Rightarrow 9S=12S-3S=12.4^{13}-12.4^{12}-(4^{12}-1)$

$\Rightarrow 9S=4^{12}.35+1$

$\Rightarrow S=\frac{4^{12}.35+1}{9}$

$\Rightarrow b=12; a=\frac{1}{9}$

$\Rightarrow 9a+b=1+12=13$

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(2A=\frac{1}{1}-\frac{1}{100}\)

\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)

Câu B và C làm tương tự.

bạn Nhi làm sai rồi

\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được

\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)

kết quả là : \(\frac{49}{100}\)