a ) 37 x 2 x 7 + 7 x 26 =
b ) 24 x 18 + 12 x 64 =
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Bài 1:
a) \(24 - (-15) - 2\)
\(=39-2\)
\(=37\)
b) \((-85) + 10 - (-85) - 50\)
\(=[(-85)-(-85)]+10-50\)
\(=0+10-50\)
\(=10-50\)
\(=-40\)
c) \(71 - (-30) - (+18) + (-30)\)
\(=[(-30)-(-30)]+71-(+18)\)
\(=0+71-18\)
\(=71-18\)
\(=53\)
d) \(-(30) - (+37) + (+37) + (-85)\)
\(=[-(+37)+(+37)]-(30)+(-85)\)
\(=0-(30)+(-85)\)
\(=(-30)+(-85)\)
\(=-115\)
e) \((35-815) - (795-65)\)
\(=(-780)-730\)
\(=-1510\)
g) \((2002-79+15) + (-79+15)\)
\(=1938+(-64)\)
\(=1874\)
Bài 2:
a) \(25 - (30+x) = x - (27-8)\)
\(25-30-x=x-27+8\)
\(x+x=25-30+27-8\)
\(2x=14\)
\(x=14\div2\)
\(x=7\)
b) \((x-12) - 15 = (20-17) - (18+x) \)
\(x-12-15=13-18-x\)
\(x-27=-5-x\)
\(x+x=-5+27\)
\(2x=22\)
\(x=22\div2\)
\(x=11\)
c) \(15 - x = 7 - (-2)\)
\(15-x=9\)
\(x=15-9\)
\(x=6\)
d) \(x - 35 = (-12) - 3\)
\(x-35=-15\)
\(x=-15+35\)
\(x=20\)
e) \(\left|5-x\right|-26=-15\)
\(\left|5-x\right|=-15+26\)
\(\left|5-x\right|=11\)
Từ đây ta có:
*Nếu \(5-x=11\)
\(x=5-11\)
\(x=-6\)
*Nếu \(5-x=-11\)
\(x=5-(-11)\)
\(x=16\)
Vậy \(x=-6;x=16\)
Tính bằng cách thuận tiện nhất :
a. 67 x 126 - 67 x 84 - 41 x 67 - 67
= 67 x 126 - 67 x 84 - 41 x 67 - 67 x 1
= 67 ( 126 - 84 - 41 - 1 )
= 67 x 0
=0
b. 36 x 2 x 2 + 7 x 28
= 2 ( 36 x 2 + 14 x 7 )
= 2 x 170
= 340
c. 24 x 18 + 12 x 64
= 12 x 2 x 18 + 12 x 64
= 12 ( 2 x 18 + 64)
= 12 x 100
= 1200
a. 67 x 126 - 67 x 84 - 41 x 67 - 67
= 67 x 126 - 67 x 84 - 41 x 67 - 67 x 1
= 67 ( 126 - 84 - 41 - 1 )
= 67 x 0
=0
b. 36 x 2 x 2 + 7 x 28
= 2 ( 36 x 2 + 14 x 7 )
= 2 x 170
= 340
c. 24 x 18 + 12 x 64
= 12 x 2 x 18 + 12 x 64
= 12 ( 2 x 18 + 64)
= 12 x 100
= 1200
a) 9 x 24 x 25
=216x25
=5400
b) 12 x 125 x 54
=1500x54
=81000
c) 64 x 125 x 875
=8000x875
=7000000
d) 425 x 7
a) 9 x 24 x 25
b) 12 x 125 x 54
c) 64 x 125 x 875
d) 425 x 7 x4 - 170 x 60
=2975x4 - 170 x 60
=11900- 170 x 60
=11900-10200
=1700
e) 8 x 9 x 14 + 6 x 7 x 12 + 19 x 4 x 18
=1008+504+1368
=2880
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
đề bài
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