`4438:7xx3`

`=634xx3`

`=1902`

__

`(1478+2354):4`

`=3832:4`

`=958`

__

`793+1608:3`

`=793+536`

`=1329`

__

`2406:3+1237`

`=802+1237`

`=2039`

bằng 10

\(A=x^5-5x^4+5x^3-5x^2+5x\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x\)

\(=x\)

\(=4\)

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

\(\frac{2}{5}-\frac{1}{7}+\frac{3}{5}.\frac{1}{3}=\frac{14}{35}-\frac{5}{35}+\frac{7}{35}=\frac{16}{35}\)

\(\frac{2}{5}-\frac{1}{7}+\frac{3}{5}\times\frac{1}{3}\)

\(=\frac{2}{5}-\frac{1}{7}+\frac{1}{5}\)

\(=\frac{9}{35}+\frac{1}{5}\)

\(=\frac{16}{35}\)

đầy đủ phần ngoặc hay gì đó

\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)

= 22 \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{110}{3}\)

\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)

= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)

= \(\dfrac{7}{12}\) 

\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5

= \(\dfrac{28}{15}\) + 5

= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)

= \(\dfrac{103}{15}\)

bạn này mới tham gia hoc24

Thanks kiuuuu

\(2\dfrac{1}{20}\)

\(\dfrac{13}{18}\)

\(\text{A}=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)

\(\frac{1}{2}.\text{A}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{100}}+\frac{100}{2^{101}}\)

\(=\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right]-\frac{100}{2^{101}}\left(\text{do}\frac{3}{2^3}=\frac{1}{2^2}+\frac{1}{2^3}\right)\)

\(=\frac{\left[1-\left(\frac{1}{2}\right)^{101}\right]}{\left(1-\frac{1}{2}\right)}-\frac{100}{2^{101}}\)

\(=\frac{\left(2^{101}-1\right)}{2^{100}}-\frac{100}{2^{101}}\)

\(\Rightarrow\text{A}=\frac{\left(2^{101}-1\right)}{2^{99}}-\frac{100}{2^{101}}\)

P/s: Sai đâu thì bn sửa nhé.

Bài này là ttoan nâng cao hả bạn

\(a,\dfrac{1}{2}:\dfrac{2}{4}:5\) -> chỉ có phép chia nên thực hiện từ trái sang phải :>

\(=1:5=\dfrac{1}{5}\) 

\(b,\dfrac{2}{5}:12:\dfrac{4}{3}\) -> tương tự câu thứ nhất :>

\(=\dfrac{1}{30}:\dfrac{4}{3}=\dfrac{1}{40}\)

1)

\(\dfrac{1}{2}:\dfrac{2}{4}:\dfrac{5}{1}=\left(\dfrac{1}{2}.\dfrac{4}{2}\right).\dfrac{1}{5}=1.\dfrac{1}{5}=\dfrac{1}{5}\)

dấu chấm là dấu nhân á .

2)

\(\dfrac{2}{5}:\dfrac{12}{1}:\dfrac{4}{3}=\left(\dfrac{2}{5}.\dfrac{1}{12}\right).\dfrac{3}{4}=\dfrac{1}{30}.\dfrac{3}{4}=\dfrac{3}{120}=\dfrac{1}{40}\)