Gọi số cần tìm có dạng là \(X=\overline{ab}\)

Khi viết thêm chữ số 0 vào giữa hai chữ số thì ta được số mới gấp 6 lần số cũ nên \(\overline{a0b}=6\cdot\overline{ab}\)

=>\(100a+b=6\left(10a+b\right)\)

=>100a+b=60a+6b

=>40a=5b

=>8a=b

=>b=8; a=1

Vậy: Số cần tìm là 18

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-20\right)=150\\ x-1+x-2+...+x-20=150\\ \left(x+x+...+x\right)-\left(1+2+...+20\right)\\ 20\cdot x-\left[\left(20-1\right):1+1\right]\cdot\left(20+1\right):2=150\\ 20\cdot x-20\cdot21:2=150\\ 20\cdot x-210=150\\ 20\cdot x=150+210\\ 20\cdot x=360\\ x=360:20\\ x=18\)

Chọn số lớn nhất phải không  em???

11: \(2^2\cdot3^2-5\cdot2\cdot3=6^2-30=36-30=6\)

12: \(3^2\cdot5-2^2\cdot7+1\cdot5=9\cdot5-4\cdot7+5\)

=45-28+5

=50-28=22

13: \(5^2\cdot2-3^2\cdot4=25\cdot2-9\cdot4=50-36=14\)

14: \(7^2\cdot3-5^2\cdot3=49\cdot3-25\cdot3=24\cdot3=72\)

15: \(2^3\cdot3^2-4^2\cdot3=8\cdot9-16\cdot3=72-48=24\)

16: \(5^2\cdot2^3+3^2\cdot7-8^2\cdot2\)

\(=25\cdot8+9\cdot7-64\cdot2\)

=200+63-128

=263-128=135

17: \(\left(5\cdot2^2-20\right):5+3^2\cdot6=\left(5\cdot4-20\right):5+9\cdot6\)

=0+54

=54

18: \(\left(24\cdot5-5^2\cdot2\right):\left(5\cdot2\right)-3\)

\(=\left(120-50\right):10-3\)

=7-3=4

19: \(\left[\left(5^2\cdot2^3-7^2\cdot2\right):2\right]\cdot6-7\cdot2^5\)

\(=\left[5^2\cdot2^2-7^2\right]\cdot6-7\cdot32\)

=(100-49)*6-224

=51*6-224

=82

20: \(\left(6\cdot5^2-13\cdot7\right)\cdot2-2^3\left(7+3\right)\)

\(=\left(6\cdot25-91\right)\cdot2-8\cdot10\)

\(=\left(150-91\right)\cdot2-80\)

=118-80=38

\(14\cdot98=14\cdot\left(100-2\right)=14\cdot100-14\cdot2\)

=1400-28

=1372

cảm ơn

\(\left(3x-16\right)\cdot49=2\cdot343\)

=>\(3x-16=2\cdot343:49\)

=>\(3x-16=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\dfrac{30}{3}=10\)

\(\left(3\cdot x-16\right)\cdot49=2\cdot343\\ \Rightarrow\left(3\cdot x-16\right)\cdot49=686\\ \Rightarrow3\cdot x-16=686:49\\ \Rightarrow3\cdot x-16=14\\ \Rightarrow3\cdot x=14+16\\ \Rightarrow3\cdot x=20\\ \Rightarrow x=\dfrac{20}{3}\)

Vậy \(x=\dfrac{20}{3}\)

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{B}+70^0+40^0=180^0\)

=>\(\widehat{B}=70^0\)

Ta có: \(\widehat{xAB}+\widehat{ABC}=70^0+110^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên Ax//BC

a) \(\dfrac{-14}{12}+0,65-\left(\dfrac{-7}{42}-0,35\right)\\ =\dfrac{-7}{6}+0,65+\dfrac{7}{42}+0,35\\ =\left(-\dfrac{7}{6}+\dfrac{7}{42}\right)+\left(0,65+0,35\right)\\ =\left(-\dfrac{7}{6}+\dfrac{1}{6}\right)+1\\ =\dfrac{-6}{6}+1=-1+1=0\)

b) \(\left(\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}\right)-\left(-\dfrac{3}{7}+1-\dfrac{13}{8}\right)\\ =\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}+\dfrac{3}{7}-1+\dfrac{13}{8}\\ =\left(\dfrac{7}{8}+\dfrac{13}{8}-\dfrac{5}{2}\right)+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-1\\ =\left(\dfrac{20}{8}-\dfrac{20}{8}\right)+\dfrac{7}{7}-1\\ =0+1-1=0\)

c) \(\dfrac{1}{2}-\dfrac{43}{101}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\\ =\dfrac{1}{2}-\dfrac{43}{101}-\dfrac{1}{3}-\dfrac{1}{6}\\ =\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)-\dfrac{43}{101}\\ =\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)-\dfrac{43}{101}\\ =0-\dfrac{43}{101}=-\dfrac{43}{101}\)

a; - \(\dfrac{14}{12}\) + 0,65 - ( - \(\dfrac{7}{42}\) - 0,35)

  = - \(\dfrac{7}{6}\) + 0,65 + \(\dfrac{7}{42}\) + 0,35

= (- \(\dfrac{7}{6}\) + \(\dfrac{7}{42}\)) + (0,65 + 0,35)

= (-\(\dfrac{49}{42}\) + \(\dfrac{7}{42}\)) + 1

= - 1 + 1

= 0