Cho A = 1,2 x ( 3,75 - 0,75 : P - 0,5 ) Tìm giá trị của P để giá trị của A = 3,6
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A = 3,6
1,2 × (3,75 - 0,75 : P - 0,5) = 3,6
3,75 - 0,75 : P - 0,5 = 3,6 : 1,2
3,75 - 0,75 : P - 0,5 = 3
3,25 - 0,75 : P = 3
0,75 : P = 3,25 - 3
0,75 : P = 0,25
P = 0,75 : 0,26
P = 3
1,2 x ( 2,5:P)=3,6
2,5:P = 3,6:1,2
2,5:P = 3
P = 2,5:3
P = \(\dfrac{5}{6}\)
a) 387*0,25 -(x: 4+x: 0,125): 0,5 = 14,25
96,75 - (x: 4 + 32x: 4): 0,5 = 14,25
(x: 4 + 32x: 4): 0,5 96,75 - 14,25
(x: 4 + 32x: 4): 0,5 = 82,5
(33x: 4): 0,5 = 82,5
33x4 = 82,5 * 0,5
33x: 441,25
33x = 41,25 * 4
b) (2,5*1,2 + x - 2,5): 2,5 = 4
2,5*1,2x-2,5 = 4*2,5
2,5*1,2+x-2,5 = 10
3+x-2,5 = 10
3 + x = 10 + 2,5
3+ x = 12,5
x = 12,5 - 3
x = 9,5
c \()\)1,2x(3,75-0,75:x-0,5)=3,6
3,75-0,75:x-0,5=3,6:1,2
3,75-0,75:x-0,5=3
0,75:x-0,5-3,75-3
0,75:x-0,5=0,75
0,75:x=0,75+0,5
0,75:x=1,25
X=0,75:1,25
x=0,6
0,75 + 1,5 \(\times\) 97,8 \(\times\) 0,5 + 0,25 \(\times\) 3 \(\times\) 1,2
= 0,75 \(\times\) 1 + (1,5 \(\times\) 0,5)\(\times\) 97,8 + (0,25 \(\times\)3) \(\times\) 1,2
= 0,75 \(\times\) 1 + 0,75 \(\times\) 97,8 + 0,75 \(\times\) 1,2
= 0,75 \(\times\) ( 1 + 97,8 + 1,2)
= 0,75 \(\times\) [1 + ( 97,8 + 1,2)]
= 0,75 \(\times\) [1 + 99]
= 0,75 \(\times\) 100
= 75
(2000×7,5+2012÷3)×(21-3,5×0,25)×(11,2-10-1,2)×(3,75-0,75)/2011+3,14=15670.6666667x20.125x0x3:2014.14=0
học tót
(2000×7,5+2012÷3)×(21-3,5×0,25)×(11,2-10-1,2)×(3,75-0,75)/2011+3,14
=(2000×7,5+2012÷3)×(21-3,5×0,25)×(1,2-1,2)×(3,75-0,75)/2011+3,14
=(2000×7,5+2012÷3)×(21-3,5×0,25)×0×(3,75-0,75)/2011+3,14
=0
a) \(210:x=2,1:0,25=8,4\)
\(x=210:8,4=25\)
b) \(x.2+x.0,5=6,25\)
\(2,5x=6,25\)
\(x=6,25:2,5=2,5\)
c) \(x:0,25-x=15,6\)
\(3x=15,6\)
\(x=5,2\)
d) \(x:12,5=3,75.4=15\)
\(x=15.12,5=187,5\)
e) \(x.12,6-x.5,6=42\)
\(7x=42\)
\(x=6\)
g) \(x:0,1-x:4-x.0,75=2,25\)
\(10x-\dfrac{x}{4}-\dfrac{3x}{4}=2,25\)
\(9x=2,25\)
\(x=0,25\)
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
a: =472x2+472x5+472x2
=472x9=4248
b: =3,6x8+7,2x4+7,2x2
=7,2(4+4+2)
=72
d: =2,4(6,2+1,9x2)=24
a: =472x2+472x5+472x2
=472x9=4248
b: =3,6x8+7,2x4+7,2x2
=7,2(4+4+2)
=72
d: =2,4(6,2+1,9x2)=24
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