a: \(1968\cdot15+1968\cdot84+1968\)

\(=1968\left(15+84+1\right)\)

\(=1968\cdot100=196800\)

b: \(6412+5378+3582+4628\)

\(=\left(6412+4628\right)+\left(5378+3582\right)\)

\(=11040+8960=20000\)

c: \(178\cdot101-178\)

\(=178\cdot101-178\cdot1\)

\(=178\left(101-1\right)=178\cdot100=17800\)

d: \(153\cdot8+153+153\)

\(=153\left(8+1+1\right)\)

\(=153\cdot10=1530\)

a) 1968 x 15 + 1968 x 84 + 1968

= 1968 x 15 + 1968 x 84 + 1968 x 1

= 1968 x (15 + 84 + 1)

= 1968 x 100 = 196800

b) 6412 + 5378 + 3582 + 4628

= (6412 + 5378) + (3582 + 4628)

= 11790 + 8210

= 20000

c) 178 x 101 - 178

= 178 x 101 - 178 x 1

= 178 x (101 - 1)

= 178 x 100

= 17800

d) 153 x 8 + 153 + 153

= 153 x 8 + 153 x 1 + 153 x 1

= 153 x (8 + 1 + 1)

= 153 x 10

= 1530

thầy cô và các bạn xin giúp em!

\(x-153-48+193=2x+193-12-153\)

\(x-48=2x-12\)

\(x=-36\)

(\(x\) \(\times\) (\(x\) + 1)): 2 = 153

(\(x\) \(\times\) (\(x\) + 1)) = 153  \(\times\) 2

\(x\) \(\times\) (\(x\) + 1) = 306

\(x\)² + \(x\) = 306

\(x^2\) + \(x\) - 306 = 0

\(x^2\) - 17\(x\) + 18\(x\) - 306 =0

\(x\) \(\times\) (\(x\) - 17) + 18 \(\times\) (\(x\) - 17) = 0

(\(x\) - 17)\(\times\) ( \(x\) + 18) = 0

\(\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\)

\(\text{#040911}\)

\(x\cdot\left(x+1\right)\div2=153\)

\(\Rightarrow x\cdot\left(x+1\right)=153\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=306\)

\(\Rightarrow x^2+x=306\)

\(\Rightarrow x^2+x-306=0\)

\(\Rightarrow x^2+18x-17x-306=0\)

\(\Rightarrow\left(x^2+18x\right)-\left(17x+306\right)=0\\ \Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\\ \Rightarrow\left(x-17\right)\left(x+18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\\ \text{Vậy, x }\in\left\{-18;17\right\}.\)

\(x\left(x+1\right):2=153\)

\(x\left(x+1\right)=306\)

\(x^2+x=306\)

\(x=17;x=-18\)

\(\text{#040911}\)

\(\left[x\cdot\left(x+1\right)\right]\div2=153\\ \Rightarrow x\cdot\left(x+1\right)=153\cdot2\\ \Rightarrow x\cdot\left(x+1\right)=306\\ \Rightarrow x^2+x=306\\ \Rightarrow x^2+x-306=0\\ \Rightarrow x^2+18x-17x-306=0\\ \Rightarrow\left(x^2+18x\right)-\left(17x+306\right)=0\\ \Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\\ \Rightarrow\left(x-17\right)\left(x+18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-17=0\\x+18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\\ \text{Vậy, x }\in\left\{-18;17\right\}.\)

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\)

   \(\Leftrightarrow\left(x+x+x+...+x\right)+\left(0+1+2+...+2019\right)=0\)( có 2020 chữ x )

   \(\Leftrightarrow2020x+2039190=0\)

   \(\Leftrightarrow x=-1009,5\)

Ta có : \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\)

\(\Rightarrow x+x+1+x+2+...+x+2019=0\)

\(\Rightarrow2020x+\left(1+2+3+...+2019\right)=0\)

\(\Rightarrow2020x+\frac{2019.2020}{2}=0\)

\(\Rightarrow2020x+2039190=0\)

\(\Rightarrow2020x=-2039190\)

\(\Rightarrow x=1009,5\)

Vậy \(x=1009,5\)

X x 3= 153                                     X : 2 = 320

X       = 153 : 3                               X       = 320 x 2

 X       = 51                                      X       = 640

\(x\cdot\left(x+1\right):2=153\)

\(\Rightarrow x\left(x+1\right)=153\cdot2\)

\(\Rightarrow x^2+x=306\)

\(\Rightarrow x^2+x-306=0\)

\(\Rightarrow x^2+18x-17x-306=0\)

\(\Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\)

\(\Rightarrow\left(x-17\right)\left(x+18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+28\right)=153\)

\(\Leftrightarrow x+1+x+2+...+x+28=153\)

\(\Leftrightarrow28x+406=153\Leftrightarrow28x=153-406=-253\)

\(\Leftrightarrow x=\dfrac{-253}{28}\)