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10) 37.(27+25)-27.(37+25)=37.27+37.25-27.37-27.25
=37.25-27.25=25(37-27)=25.10=250
11)3.(-5+3)+7(2+5)=3.(-2)+7.7=-6+49=43
12)-3.(2-7)-6(8-5)=-3.(-5)-6.3=15-18=-3
1) 9-25=(7-x)-(25+7)
(7-x)-(25+7)=9-25
(7-x)-(25+7)=-16
(7-x)-32=-16
7-x=-16+32
7-x=16
x=7-16
x=-9
2) (27-514) -(486-73)+x=7
(27-514) -(486-73)+x=7
-487-413+x=7
-1011+x=7
x=7-(-1011)
x=7+1011
x=1018
3) 25+5+37-25+6-29-x=37
25+5+37-25+6-29-x=37
67-25+6-29-x=37
42+6-29-x=37
48-29-x=37
19-x=37
x=19-37
x=-18
4) 14+(-12)+x =10-/-15/+ /-3/
14+(-12)+x =10-15+3
14+(-12)+x =-5+3
14+(-12)+x =-2
2+x=-2
x=-2-2
x=-4
a) \(17-\left(-9\right)-27-19\\ =17+9-27-19\\ =\left(17-27\right)+\left(9-19\right)\\ =\left(-10\right)+\left(-10\right)=-20\)
b) \(3\cdot\left(-9\right)+9\\ =\left(-3\right)\cdot9+9\\ =9\left(-3+1\right)\\ =9\cdot\left(-2\right)=-18\)
c) \(-\left(-24\right)+\left(-37\right)-25+14\\ =24-37-25+14\\ =\left(24+14\right)-\left(37+25\right)\\ =38-62=-24\)
d) \(\left(-7\right)\cdot5-9\cdot\left(-4\right)-105\\ =\left(-7\right)\cdot5-9\cdot\left(-4\right)-21\cdot5\\ =5\left(-7-21\right)-\left(-36\right)\\ =5\cdot\left(-28\right)+36\\ =-140+36=-104\)
e) \(\left(-7\right)^2+32\div\left(-8\right)\\ =49+\left(-4\right)\\ =45\)
f) \(\left(-2\right)^2-36\div\left(-6\right)-\left|-7\right|\\ =4-\left(-6\right)-7\\ =4+6-7\\ =10-7=3\)
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 ...
a) \(\dfrac{7}{-25}+\dfrac{8}{25}=\dfrac{-7}{25}+\dfrac{8}{25}=\dfrac{1}{25}\)
\(\dfrac{4}{5}+\dfrac{4}{-18}=\dfrac{4}{5+\left(-18\right)}=\dfrac{4}{-13}=\dfrac{-4}{13}\)
\(\dfrac{7}{21}+\dfrac{9}{-36}=\dfrac{1}{3}+\dfrac{1}{-4}=\dfrac{1}{3+\left(-4\right)}=\dfrac{1}{-1}=1\)
b) \(\dfrac{12}{20}-\dfrac{2}{5}=\dfrac{12}{20}+\left(\dfrac{-2}{5}\right)=\dfrac{3}{5}+\dfrac{-2}{5}=\dfrac{3+\left(-2\right)}{5}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-\dfrac{-5}{6}=\dfrac{2}{3}+\dfrac{5}{6}=\dfrac{4}{6}+\dfrac{5}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(6-\dfrac{3}{20}=6+\dfrac{-3}{20}=\dfrac{60}{60}+\dfrac{-9}{60}=\dfrac{51}{60}\)
c) \(\dfrac{2}{21}.\dfrac{-7}{3}=\dfrac{2.\left(-1\right)}{7.3}=\dfrac{-2}{21}\)
\(\dfrac{27}{28}.\left(-21\right)=\dfrac{27.\left(-21\right)}{28}=\dfrac{-567}{28}\)
\(\dfrac{-15}{7}.\dfrac{-14}{25}=\dfrac{-3.\left(-2\right)}{1.5}=\dfrac{6}{5}\)
a) (135 - 35) . (-37) + 37 . (-42 - 58)
= 100 . 0 . (-100)
= 0
b) -65 . (87 - 17) - 87 . (17 - 65)
= -65 . 70 - 87 . (-48)
= -4550 - 135
= -4415
cũng dễ mà
mk ghi kết quả thui ha
a) = 250
b) = -27
c) = -3
tick nhé