\(48\times42+32\times8+5\times6\)

\(=16\times3\times42+16\times2\times8+16\times5\)

\(=16\times\left(126+16+5\right)\)

\(=16\times147\)

\(=2352\)

\(a,\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)

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

\(=>x=\dfrac{8}{15}\)

\(b,x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)

\(=>x.\dfrac{10}{3}=\dfrac{10}{3}:\dfrac{17}{4}\)

\(=>x=\dfrac{4}{17}\)

\(c,5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(=>\dfrac{17}{3}:x=\dfrac{7}{6}\)

\(=>x=\dfrac{34}{7}\)

`B19:`

\(a,A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

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

\(=>A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(b,B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=>B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=>B=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(c,C=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{2023}\right)\left(1-\dfrac{1}{2024}\right)\)

\(=>C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{2022}{2023}.\dfrac{2023}{2024}\)

\(=>C=\dfrac{1}{2024}\)

\(d,D=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=>D=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right):\dfrac{7}{4}\)

\(=>D=\dfrac{59}{15}-\dfrac{35}{6}:\dfrac{7}{4}\)

\(=>D=\dfrac{59}{15}-\dfrac{10}{3}=\dfrac{3}{5}\)

Số học sinh giỏi của lớp 7A là:

\(45\times20\%=9\left(hs\right)\)

Số học sinh trung bình của lớp 7A là:

\(\left(45-9\right)\times\dfrac{1}{3}=12\left(hs\right)\)

Số học sinh khá của lớp 7A là:

\(45-9-12=24\left(hs\right)\)

\(m_{M_2O_3}=2.m_M+3.m_O=160\left(amu\right)\)

\(=>2.m_M+48=160\)

\(=>2.m_M=112\)

\(=>m_M=56\left(amu\right)\)

`=>` NTK của `M` là `56amu`

`M` là `Iron(Fe)`

Thời gian đi hết qđ AB:

\(t_1=\dfrac{s_1}{v_1}=\dfrac{2}{40}=0,05\left(h\right)=3min\)

Thời gian đi hết qđ BC:

\(t_2=\dfrac{s_2}{v_2}=\dfrac{3}{60}=0,05\left(h\right)=3min\)

Thời gian đi hết qđ CD:

\(t_3=\dfrac{s_3}{v_3}=\dfrac{5}{50}=0,1\left(h\right)=6min\)

Thời gian đi hết 3 qđ:

\(t=t_1+t_2+t_3=3+3+6=12min\)

`=>` tg đi hết 3 qđ là `12` phút

đổi: `8`m/s=`28,8`km/h

Thời gian đi hết đoạn đường từ A->B :

\(t=\dfrac{s}{v}=\dfrac{4}{28,8}=0,14\left(h\right)=8min24s\)

1. a.about          b.above         c.tonight          d.ticket

2. a.before         b.study          c.little              d.people

3. a.between      b.website      c.second         d.never

4. a.hungry         b.minute       c.address        d.movie

5. a.email            b.goodbye    c.father           d.happy

6. a.bedroom      b.kitchen       c.number        d.enjoy

7. a.begin            b.better         c.player          d.doctor

8. a.teacher         b.picture       c.behind         d.mother

9. a.below           b.bathroom   c.body            d.city

10. a.nation         b.reason       c.thoughtful    d.provide

\(A=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0=>A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)      \(\left(đpcm\right)\)

\(a.\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

\(=>\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{x-2y+3z-6}{8}=\dfrac{14-6}{8}=1\)

\(=>\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.=>\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.=>\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)

\(b,\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=>\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=\dfrac{2x+3y-z-5}{9}=\dfrac{50-5}{9}=5\)

\(=>\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.=>\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.=>\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)