\(12\cdot4^3+16\cdot2\\ =12\cdot4^3+4^2\cdot2\\ =4^2\cdot\left(12\cdot4+2\right)\\ =4^2\cdot\left(48+2\right)\\ =4^2\cdot50\\ =16\cdot50\\ =800\)  

$17+x-(352-400)=-32$

$\Rightarrow x-(-48)=-32-17$

$\Rightarrow x+48=-49$

$\Rightarrow  x=-49-48$

$\Rightarrow x=-97$

\(\left[100:\left(3^3-2\right)\right]-4\)

\(=\left[100:\left(27-2\right)\right]-4\)

\(=\left(100:25\right)-4\)

\(=4-4\)

\(=0\)

\(\left[100:\left(3^3-2\right)\right]-4\)

\(=\left[100:\left(27-2\right)\right]-4\)

\(=\left[100:25\right]-4\)

\(=4-4\)

\(=0\)

-39 nhé

\(-15+\left(-24\right)=-39\)

1: \(\dfrac{3}{4}:\dfrac{1}{2}+x=\dfrac{2}{3}\)

=>\(x+\dfrac{3}{4}\cdot2=\dfrac{2}{3}\)

=>\(x+\dfrac{3}{2}=\dfrac{2}{3}\)

=>\(x=\dfrac{2}{3}-\dfrac{3}{2}=\dfrac{4}{6}-\dfrac{9}{6}=-\dfrac{5}{6}\)

2: \(\dfrac{7}{4}+\dfrac{1}{4}:x=2\)

=>\(\dfrac{1}{4}:x=2-\dfrac{7}{4}=\dfrac{1}{4}\)

=>\(x=\dfrac{1}{4}:\dfrac{1}{4}=1\)

3: \(\dfrac{48}{64}:\dfrac{12}{16}+0,25=\dfrac{3}{4}:\dfrac{3}{4}+0,25=1+0,25=1,25\)

4: \(\dfrac{3}{4}\cdot\dfrac{6}{9}+\dfrac{7}{12}\cdot6=\dfrac{18}{36}+\dfrac{7}{2}=\dfrac{1}{2}+\dfrac{7}{2}=\dfrac{8}{2}=4\)

5: \(5\cdot\dfrac{x}{6}-\dfrac{1}{4}=\dfrac{7}{2}\)

=>\(\dfrac{5}{6}x=\dfrac{7}{2}+\dfrac{1}{4}=\dfrac{15}{4}\)

=>\(x=\dfrac{15}{4}:\dfrac{5}{6}=\dfrac{15}{4}\cdot\dfrac{6}{5}=\dfrac{3}{2}\cdot3=\dfrac{9}{2}\)

6: \(\dfrac{3}{x+1}-\dfrac{1}{4}=\dfrac{7}{4}\)(ĐKXĐ: x<>-1)

=>\(\dfrac{3}{x+1}=\dfrac{7}{4}+\dfrac{1}{4}=\dfrac{8}{4}=2\)

=>\(x+1=\dfrac{3}{2}\)

=>\(x=\dfrac{1}{2}\left(nhận\right)\)

\(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{103\cdot105}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{103}-\dfrac{1}{105}\)

\(=\dfrac{1}{3}-\dfrac{1}{105}=\dfrac{34}{105}\)

\(P=\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{103\times105}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{103}-\dfrac{1}{105}\)

\(=\dfrac{1}{3}-\dfrac{1}{105}=\dfrac{34}{105}\)

Công thức: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)

a) Số phần tử của tập hợp C là:

\(\left(296-2\right):3+1=99\) (phần tử)

b) Số phần tử của tập hợp D là:

\(\left(283-7\right):4+1=70\) (phần tử)

Đề bị lỗi rồi em nhé, chưa đầy đủ em ơi!

Đề lỗi 

gọi x là số tem loại A; y là số tem loại B

tổng số tiền mà anh đã mua tem là:

10 000 - 200 = 9 800 (VNĐ)

theo đề ta có: 1300x + 700y = 9800

13x + 7y = 98

để thoả mãn đề bài thì các giá trị của x và y là số nguyên

13x + 7 × 1 = 98

13x + 7 = 98

13x = 91

x = 7

y = \(\dfrac{98-13\cdot7}{7}=1\)

tổng số tem đã mua là: 7 + 1 = 8 (tem)

vậy số tem đã mua là 8

a) \(-\dfrac{3}{7}-x=-\dfrac{1}{2}\\ x=-\dfrac{3}{7}-\left(-\dfrac{1}{2}\right)\\ x=\dfrac{-3}{7}+\dfrac{1}{2}=\dfrac{-6}{14}+\dfrac{7}{14}=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\\ x-\dfrac{4}{5}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\\ x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5}{30}+\dfrac{24}{30}\\ x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=-\dfrac{8}{11}\\ -x=-\dfrac{8}{11}+\dfrac{3}{4}\\ -x=-\dfrac{32}{44}+\dfrac{33}{44}=\dfrac{1}{44}\\ x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ \dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\ x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\\ x=\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}\\ x=-\dfrac{9}{60}\)

a) \(\dfrac{-3}{7}-x=\dfrac{-1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{7}\)

\(\Rightarrow x=\dfrac{1}{14}\)

Vậy \(x=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{4}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}+\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{29}{30}\)

Vậy \(x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(\Rightarrow-x=\dfrac{3}{4}-\dfrac{8}{11}\)

\(\Rightarrow-x=\dfrac{1}{44}\)

\(\Rightarrow x-\dfrac{1}{44}\)

Vậy \(x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{-3}{20}\)

Vậy \(x=\dfrac{-3}{20}\)