( 15 x 28 - 5 ) : = 200 : 0.4
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(15 x 28 - x) : 2/5 = 200 : 0,4
=> (420 - x) : 2/5 = 500
=> 420 - x = 500 x 2/5
=> 420 - x = 200
=> x = 420- 200
=> x = 220
(x+74) - 318=200
x+74 =200+318
x+74 =518
x =518-74
x = 444
[ (x-29) - 11 ] : 2 = 8
[ (x-29) - 11 ] = 8x2
[ (x-29) - 11 ] = 16
x-29 = 16 + 11
x-29 = 27
x = 27 + 29
x = 56
3636 : (12x -91) =36
12x-91 =3636 : 36
12x-91 = 101
12x = 101 + 91
12x = 192
x = 192 : 12
x = 16
[ ( 250 - 25 ) : 15 ] : x = ( 450 - 60 ) : 130
[ 225 : 15 ] : x = 390 : 130
15 : x = 3
x = 15 : 3
x = 5
[ ( 4x + 28 ) . 3 - 55 ] : 5 = 35
[ ( 4x + 28 ) . 3 - 55 ] = 35 : 5
[ ( 4x + 28 ) . 3 - 55 ] = 7
( 4x + 28 ) . 3 = 7 + 55
( 4x + 28 ) . 3 = 62
( 4x + 28 ) = 62 : 3
SAI ĐỀ RỒI TÍNH KO RA
Bài 2:
a, 7,2: 2,4.\(x\) = 45
3.\(x\) = 45
\(x\) = 45: 3
\(x\) = 15
b, 9,15.\(x\) + 2,85.\(x\) = 48
\(x\).( 9,15 + 2,85) = 48
12\(x\) = 48
\(x\) = 48: 12
\(x\) = 4
c) (x.3 + 4) : 5 = 8
x.3 + 4 = 8 x 5
x.3 + 4 = 40
x.3 = 40 - 4
x.3 = 36
x = 36 : 3
x = 12
45^10*5^20/75^15
=5^10*9^10*5^20/(5^2)^15
=5^10*5^20*9^10/5^30
=9^10
(0.8)^5/(0.4)^6
=(0.4)^5*2^5/(0.4)^6
=2^5/(0.4)
=32/(0.4)
=80
2^15*9^4/6^6*8^3
=2^15*(3^2)^4/2^6*3^6*(2^3)^3
=2^15*3^8/2^6*3^6*2^9
=3^2
=9
(4510 x 520 ) / 7515=243
(0.8)5 / (0.4)6=0
215 x 94 / 66 x 83=9
\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}=\frac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}=\frac{3^{20}.5^{30}}{3^{15}.5^{30}}=3^5=243\)
\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4.2\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}=\frac{2^5}{0,4}=\frac{32}{\frac{2}{5}}=32.\frac{5}{2}=80\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
a: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)
Do đó: BCNN(12;25;30)\(=2^2\cdot3\cdot5^2=300\)
x⋮12; x⋮25; x⋮30
=>x∈BC(12;25;30)
=>x∈B(300)
mà 0<x<=500
nên x=300
b: \(12=2^2\cdot3;21=3\cdot7;28=2^2\cdot7\)
Do đó: BCNN(12;21;28)\(=2^2\cdot3\cdot7=4\cdot3\cdot7=84\)
x⋮12; x⋮21; x⋮28
=>x∈BC(12;21;28)
=>x∈B(84)
mà 150<x<300
nên x∈{168;252}
c: \(15=3\cdot5;12=2^2\cdot3;18=2\cdot3^2\)
Do đó: BCNN(15;12;18)\(=2^2\cdot3^2\cdot5=36\cdot5=180\)
x⋮15; x⋮12; x⋮18
=>x∈BC(15;12;18)
=>x∈B(180)
mà 0<x<300
nên x=180
d: \(6=2\cdot3;8=2^3;12=2^2\cdot3\)
Do đó: BCNN(6;8;12)\(=2^3\cdot3=24\)
x⋮6; x⋮8; x⋮12
=>x∈BC(6;8;12)
mà x nhỏ nhất và x<>0
nên x=BCNN(6;8;12)
=>x=24
e: \(10=2\cdot5;12=2^2\cdot3;60=2^2\cdot3\cdot5\)
Do đó: BCNN(10;12;60)\(=2^2\cdot3\cdot5=60\)
x⋮10; x⋮12; x⋮60
=>x∈BC(10;12;60)
=>x∈B(60)
mà 120<=x<200
nên x∈{120;180}
f: x+10⋮5
mà 10⋮5
nên x⋮5(1)
x-18⋮6
18⋮6
Do đó: x⋮6(2)
x+21⋮7
mà 21⋮7
nên x⋮7(3)
Từ (1),(2),(3) suy ra x∈BC(5;6;7)
=>x∈B(210)
mà 500<x<700
nên x=630
g: x chia 5 dư 3
=>x-3⋮5
=>x-3+5⋮5
=>x+2⋮5(4)
x chia 6 dư 4
=>x-4⋮6
=>x-4+6⋮6
=>x+2⋮6(5)
Từ (4),(5) suy ra x+2∈BC(5;6)
=>x+2∈B(30)
=>x+2∈{30;60;90;...}
=>x∈{28;58;88;...}
mà x<59
nên x∈{28;58}
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
\(\left(15\times28-5\right):x=200:0,4\)
\(415:x=500\)
\(x=\frac{415}{500}=\frac{83}{100}=0,83\)