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

1. helps

2. eat

3. washes

1 helps 

2 eat

3 washes

Sally got out of bed, opened the window and went into the bedroom. then she has breakfast.

gets - opens - goes - has

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

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

$\Rightarrow x+48=-49$

$\Rightarrow  x=-49-48$

$\Rightarrow x=-97$

1975

Chiến tranh Việt Nam kết thúc vào lúc 11 giờ 30 phút trưa , ngày 30 tháng 4 năm 1975.

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

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

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

\(=4-4\)

\(=0\)

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

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

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

\(=4-4\)

\(=0\)

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