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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
a) |-18| . 941 + 59 . 18
= 18 . 941 + 59 . 18
= ( 941 + 59 ) . 18
= 1000 . 18
= 18000
b) 81 : 33 + 16 : 23
= 81 : 27 + 16 : 8
= 3 + 2
= 5
c) 30 - [ 40 - ( 6 - 1)2]
= 30 - [ 40 - 52 ]
= 30 - [ 40 - 25 ]
= 30 - 15
= 15
d) 17 . 85 + 15 . 17 - 150
= ( 85 + 15 ) . 17 - 150
= 100 . 17 - 150
= 1700 - 150
= 1550
a, \(\left|-18\right|\cdot941+59\cdot18=18\cdot941+59\cdot18=18\cdot\left(941+59\right)=18\cdot1000=18000\)
b, \(81:3^3+16:2^3=81\cdot27+16\cdot8=2187+128=2315\)
c, \(30-\left[40-\left(6-1\right)^2\right]=30-\left[40-5^2\right]=30-\left[40-25\right]=30-15=15\)
d, \(17\cdot85+15\cdot17-150=17\cdot\left(85+15\right)-150=17\cdot100-150=1700-150=1550\)
\(a,\left(4+32+6\right)+\left(10-32-2\right)\\ =\left(10+32\right)+\left(10-32-2\right)\\ =10+32+10-32-2\\ =20-2\\ =18\\ b,300:4+300:6-25\\ =300:\left(4+6\right)-25\\ =300:10-25\\ =30-25\\ =5\\ c,17.\left[29-\left(-111\right)\right]+29.\left(-17\right)\\ =17.\left(29+111\right)+29.\left(-1\right).17\\ =17.\left(29+111-29\right)\\ =17.111=1887\\ d,19.43+\left(-20\right).43-\left(-40\right)\\ =19.43+\left(-20\right).43+40\\ =43.\left(19-20\right)+40\\ =-43+40\\ =-3\)
Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right).\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot\left(\dfrac{3-2-1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot\dfrac{0}{6}\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot0=0\)