a, 15(21/43+22/43)=15.1=15

b, 399(45+55) / 1995(1996-1991)

=399.100 / 1995.5

=4

=0

đơn giản vậy

1996 x 1995 - 996 - 1000 - 1996 x 1994

= 1996 x (1995 - 1994) - (996 + 1000)

= 1996 - 1996

= 0

1996 x 1995 - 996 - 1000 - 1996 x 1994

= 1996 x 1995 - ( 1000 + 996 ) - 1996 x 1994

= 1996 x 1995 - 1996 - 1996 x 1994

= 1996 x ( 1995 - 1 - 1994 )

= 1996 x 0

= 0

X thôi nha

a, x = 40 000 + 8 000 + 700 + 50 = 48 750

b, x = 600 000 + 5 000 + 40 = 605 040

\(5.\left(x+35\right)=515\)

     \(x+35=515:5\)

     \(x+35=103\)

     \(x=103-35\)

 \(\Rightarrow x=68\)

\(\left(x+40\right).15=75.12\)

\(\Rightarrow\left(x+40\right)=5.12\)[ rút gọn theo t/c đẳng thức ]

\(x+40=60\)

\(x=20\)

trả lời giùm mình mấy câu kia với

a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có nghiệm là x = 2004 .

b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)

=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)

=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)

=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình có nghiệm là x = 100 .

a) 399(45+55) / 1995(1996-1991)

=39900 / 9975

=4

b) 

a) \(\frac{399\cdot45+55\cdot399}{1995\cdot1996-1991\cdot1995}\)

\(=\frac{399\cdot\left(45+55\right)}{1995\cdot\left(1996-1991\right)}\)

\(=\frac{399\cdot100}{1995\cdot5}\)

\(=4\)

=))

\(a,\frac{399×\left(45+55\right)}{1995×\left(1996-1991\right)}\)

\(=\frac{39900}{9975}\)

\(=4\)

a) 399(45+55) / 1995(1996-1991)

=39900 / 9975

=4

a) \(\frac{399\cdot45+55\cdot399}{1995\cdot1996-1991\cdot1995}\)

\(=\frac{399\cdot\left(45+55\right)}{1995\cdot\left(1996-1991\right)}\)

\(=\frac{399\cdot100}{1995\cdot5}\)

\(=4\)

=))

b) 1000.1995+1995.996-996

    1000+1000.1994+1994.996

=1000.1995+996.1994

   1000.1995+996.1994

=1

1.53x201=10653

2.

a)4 xX-15x4=6x15+X

<=>4x-60=90+x

<=>4x-x=60+90

<=>3x=150

<=>x=50

b(102+28-2x):20-5=1

<=>(130-2x):20=6

<=>130-2x=120

<=>2x=10

<=>x=5

c)1999x-x=1999+1997+1999

<=>1998x=5995

<=>\(x\approx3\)

d)105-5x=150:15

<=>105-5x=10

<=>5x=95

<=>x=19

e)(x+1)+(x+2)+(x+3)+...+(x+6)=78

<=>6x+21=78

<=>6x=57

<=>x=9,5

1) 53x201=53x200+53=10600+53=10653

2a)

4x-15x4=6x15+x

4x-60=90+x

4x-x=90+60

3x=150

x=50

b) x=5

c) x=3

d)x=19

e) x=9,5

các bạn ơi 4x và 3x là j vậy ??