Nếu x=15 thì 85+x=
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3 x 15 + 21 x 15 + 85 x 5
= 45 + 315 + 425
= 785
15 - 30 + 40
= 25
21 + 19 - 50 + 10
= 0
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=-\dfrac{1}{20}+2\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{12}\times\dfrac{3}{12}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=-\dfrac{9}{20}\)
\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)
\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)
\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)
\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)
\(=\dfrac{44}{20}-\dfrac{5}{20}\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)
\(=\dfrac{6}{20}-\dfrac{15}{20}\)
\(=-\dfrac{9}{20}\)
a) 327 x 15 + 327 x 85
= 327 x (15+85)
= 327 x 100
= 32700
b) 405 x 250 + 750 x 405
= 405 x (250+750)
= 405 x 1000
- 405000
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 \(x=-6;x=16\)
ta có:(x+y)^2=1 (vì x+y=1)
=> x^2+2xy+y^2=1
=>x^2+y^2=1-2xy
=>1-2xy=85 (vì x^2+y^2=85)
=>2xy=-84
=>xy=-42 (1)
Ta có: x^3+y^3=(x^3+3x^2y+3xy^2+y^3)-3x^2y+3xy^2 (thêm bớt 3x^2y và 3xy^2)
=(x+y)^3-3xy(x+y)
=1-3xy (2)
Từ (1) va (2) =>x^3+y^3=1-(-126)=127
Vậy x^3+y^3=127
Ta có: \(\dfrac{x-25}{75}+\dfrac{x-15}{85}+\dfrac{x-5}{95}+\dfrac{x-145}{15}=0\)
\(\Leftrightarrow\dfrac{x-25}{75}-1+\dfrac{x-15}{85}-1+\dfrac{x-5}{95}-1+\dfrac{x-145}{15}+3=0\)
\(\Leftrightarrow\dfrac{x-100}{75}+\dfrac{x-100}{85}+\dfrac{x-100}{95}+\dfrac{x-100}{15}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}\right)=0\)
mà \(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
a) 15 x 141 + 59 x 15
= 15.(141 + 59)
= 15.200
= 3000
b) 17 x 85 + 15 x 17 - 120
= 17(85 + 15) - 120
= 17.100 - 120
= 1700 - 120
= 1580
15*(141+59)
=15*200
=3000
17*(85+150)-120
=17*100-120
=1700-120
=1580
a: 15/x=10/15
nên x=22,5
b: 3/x=51/85
=>3/x=3/5
hay x=5
Nếu x = 15 thì ta có : 85 + x = 85 + 15 = 100
thì x =100 nha