trả lời trước 1h hộ mình nhé ình vội lắm rồi
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\(14\cdot98=14\cdot\left(100-2\right)=14\cdot100-14\cdot2\)
=1400-28
=1372
\(\left(3x-16\right)\cdot49=2\cdot343\)
=>\(3x-16=2\cdot343:49\)
=>\(3x-16=2\cdot7=14\)
=>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
\(a,\left(-\dfrac{13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)\\ =-\dfrac{13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =-\dfrac{3}{7}.\)
\(\dfrac{2}{5}-x=2-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}-x=\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{17}{20}.\)
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