\(|x|-\frac{1}{2}+2|x|=\frac{3}{4}\)

\(3|x|-\frac{1}{2}=\frac{3}{4}\)

\(3|x|=\frac{1}{4}\)

\(|x|=\frac{1}{4}\div3\)

\(|x|=\frac{1}{12}\)

\(\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{1}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{12};-\frac{1}{12}\right\}\)

cảm ơn bạn nhé

d) B(23) = {0; 23; 46; 69; 92; ...}

     B(10) = {0; 10; 20; 30; 40; ...}

     B(8) = {0; 8; 16; 32; 64; ...}

e) B(3) = {0; 3; 6; 9; 12; ...}

     B(12) = {0; 12; 24; 36; 48; ...}

     B(9) = {0; 9; 18; 27; 36; ...}

g) B(18) = {0; 18; 36; 54; 72; ...}

     B(20) = {0; 20; 40; 60; 80; ...}

     B(14) = {0; 14; 28; 32; 46; ...}

giúp tui với help me:<

muốn giúp nhưng cs tội là ko bt thoi -_-

https://olm.vn/hoi-dap/tim-kiem?q=H%E1%BB%8Dc+k%E1%BB%B3+1,+s%E1%BB%91+h%E1%BB%8Dc+sinh+gi%E1%BB%8Fi+c%E1%BB%A7a+l%E1%BB%9Bp+6D+b%E1%BA%B1ng+2/7+s%E1%BB%91+h%E1%BB%8Dc+sinh+c%C3%B2n+l%E1%BA%A1i.+Sang+h%E1%BB%8Dc+k%C3%AC+2,+s%E1%BB%91+h%E1%BB%8Dc+sinh+gi%E1%BB%8Fi+t%C4%83ng+th%C3%AAm+8+b%E1%BA%A1n+(s%E1%BB%91+h%E1%BB%8Dc+sinh+c%E1%BA%A3+l%E1%BB%9Bp+ko+%C4%91%E1%BB%95i),+n%C3%AAn+s%E1%BB%91+h%E1%BB%8Dc+sinh+gi%E1%BB%8Fi+b%E1%BA%B1ng+2/3+s%E1%BB%91+c%C3%B2n+l%E1%BA%A1i.+H%E1%BB%8Fi+h%E1%BB%8Dc+k%C3%AC+1+l%E1%BB%9Bp+6D+c%C3%B3+bao+nhi%C3%AAu+h%E1%BB%8Dc+sinh+gi%E1%BB%8Fi?&id=82053

Ta có: 

\(n^2=n^2-n+n=n\left(n-1\right)+n\)

Áp dụng ta được: 

\(A=1^2+2^2+3^2+...+100^2\)

\(=1.2+2.3+...+99.100+1+2+3+...+100\)

Đặt \(B=1.2+2.3+...+99.100\)

\(C=1+2+3+...+100\)

\(3B=1.2.3+2.3.3+...+99.100.3\)

\(=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(=99.100.101\)

\(B=\frac{99.100.101}{3}=100.101.33\)

\(C=1+2+...+100=\frac{100.101}{2}\)

\(A=B+C=100.101.33+\frac{100.101}{2}=\frac{100.101.\left(200+1\right)}{6}\)

\(A=\frac{10^{2019}+1}{10^{2020}+10}=\frac{10^{2019}+1}{10\left(10^{2019}+1\right)}=\frac{1}{10}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=1\\b=10\end{cases}}\)

\(B=\frac{10^{2018}+1}{10^{2020}+100}=\frac{10^{2018}+1}{100\left(10^{2018}+1\right)}=\frac{1}{100}=\frac{c}{d}\Rightarrow\hept{\begin{cases}c=1\\d=100\end{cases}}\)

\(\Rightarrow ad+bc=1.100+10.1=110\)

\(c+3a=2016\Leftrightarrow a=\frac{2016-c}{3}\ge0\Leftrightarrow c\le2016\)

\(c+2b=2018\Leftrightarrow b=\frac{2018-c}{2}\Leftrightarrow c\le2018\)

\(a+b+c=\frac{2016-c}{3}+\frac{2018-c}{2}+c\)

\(=\frac{2016}{3}+\frac{2018}{2}+\frac{c}{6}\le\frac{2016}{3}+1009+\frac{2016}{6}=2017\)

Dấu \(=\)khi \(c=2016\Rightarrow a=0,b=1\).

24;x=8

20;x=5

35;x=7

Học tốt!!!

Vì 24 chia hết cho x, 20 chia hết cho x, 35 chia hết cho x và x lớn nhất nên :

x là ƯCLN của 24 ; 20 và 35

Ta có : 

24 = 2³ x 3

20 = 2² x 5

35 = 5 x 7

=> ƯCLN( 24;24;35 ) = 1

Vậy x = 1

Sai thì xl nha TvT