\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\)\(:16\frac{2}{3}=0\)

\(\Rightarrow\)\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0\)

\(\Rightarrow\)\(5\frac{3}{8}+x-7\frac{5}{24}=2\)

\(\Rightarrow5\frac{3}{8}+x=9\frac{5}{24}\)

\(\Rightarrow x=3\frac{5}{6}\)

( Cách làm thì đúng rồi nhưng đáp án tớ ko chắc chắn )

1/(-37) +14+26+37

=[(-37)+37] +(14+26)

=       0       +   40

=    40

2/(-24) +6 +10+24

=[(-24)+24] +(10+6)

=     0         +16

=16

3/15+23+9+(-25) +9+(-23)

=[(-23)+23] +[(-25)+15] +(9+9)

=     0         +       (-10)    +18

=     0         +           8

=   8

4/60 +33 +(-50) +(-33)

 =[(-33)+33]+[60 +(-50)]

=    0          +   10

=    10

5/(-16)+ (-209) + (-14) +209

=[(-209)+209] +[(-16)+(-14)]

=     0              +(-30)

=      -30

6/(-12) +(-13)+36 +(-11)

  =[(-12) +(-11)] +[ 36+(-13)]

 =        (-23)       +       23

=     0

 Chúc bn hok tốt !!!

1, (-37)+14+26+37=[(-37)+37]+14+26=0+40=40

2, (-24)+6+10+24=[(-24)+24]+6+10=0+16=16

3 , 15+23+(-25)+(-23)=[15+(-25)]+[23+(-23)]=-10

4, 60+33+(-50)+(-33)= [60+(-50)]+[33+(-33)]=10

5, (-16)+(-209)+(-14)+209=[(-16)+(-14)]+[(-209)+209]=-30

6, (-12)+(-13)+36+(-11)=[(-12)+(-11)]+[(-13)+36]=(-23)+23=0

Đề bài: Tìm x

Ta có : \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)

\(3x+6=-4x+20\)

\(3x+4x=-6+20\)

\(7x=14\)

\(x=14\div7\)

\(x=2\)

Vậy \(x=2\).

\(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

  3/x-5= - 4/x+2

 =>3.(x+2)=-( - 4).(x-5)

=> 3x+3.2= - 4x-(-20)

=>3x+6= - 4x+20

=>4x+3x=20-6

=> 7x=14

=>x=14:7

=>x=2

a) x = 1

b) x = 2

c) ko có Đ/Á

d) x = 3

a) 5 . x³ - 5 = 0

=> 5 . ( x³ - 1 ) = 0

=> x³ - 1 = 0

=> x³ = 1

=> x³ = 1³

=> x = 1

b) ( x + 1 )³ = 27

=> ( x  +1 )³ = 3³

=> x + 1 = 3

=> x = 2

c) Sửa đề :

( x + 1 )⁴ = 16

=> ( x + 1 )⁴ = 16

=> ( x + 1 )⁴ = 2⁴

=> x + 1 = 4

=> x = 2

h) ( 2x - 1 )³ = 125

=> ( 2x - 1 )³ = 5³

=> 2x - 1 = 5

=> 2x = 6

=> x = 3

a) 5x^3 =5     b)x+1=3     h) 2x-1 = 5

<=>x^3=1     <=>x=2      <=> 2x=6

<=>x=1                          <=> x=3

Bạn xem lại đề câu c mình làm dc nhưng số ko đẹp

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)