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Ta có: \(1-\left(\frac{49}{9}+x-\frac{133}{18}\right):\frac{11}{2}=\frac{4}{9}\)
\(\Rightarrow\left(\frac{49}{9}+x-\frac{133}{18}\right):\frac{11}{2}=1-\frac{4}{9}=\frac{5}{9}\)
\(\frac{49}{8}+x-\frac{133}{19}=\frac{5}{9}.\frac{11}{2}=\frac{55}{18}\)
\(\frac{49}{8}+x=\frac{55}{18}+\frac{133}{19}=\frac{181}{18}\)
\(\Rightarrow x=\frac{181}{18}-\frac{49}{8}=\frac{283}{72}\)
Vậy \(x=\frac{283}{72}\)
1 - ( 49/9 + x -133/18 ):11/2 = 4/9
1 - ( 49/9 + x - 133/18 ) = 4/9 . 11/2
1 - ( 49/9 + x - 133/18 ) = 22/9
49/9 + x - 133/18 = 1 - 22/9
49/9 + x - 133/18 = - 13/9
49/9 + x = -13/9 - 133/18
49/9 + x = -53/6
x = -53/6 -49/9
x = -257/18
b) 541 + (218 - x) = 678
218-x=678-541
218-x=137
x=181
(139139 . 133 - 133133 . 139) : (2 + 4+ 6 + ... + 2002)
= (139 . 1001 . 133 - 133 . 1001 . 139) : (2 + 4 + 6 + ... + 2002)
= 0 : (2 + 4 + 6 + ... + 2002)
= 0
a: \(A=\left[111\cdot\left(38\cdot3-133\right)\right]\cdot2018\)
\(=111\cdot\left(-19\right)\cdot2018=-4255962\)
b: \(B=\left[25\left(1+3\cdot3-5\cdot2\right)\right]:20182019\)
\(=25\left(1+9-10\right):20182019=0\)
c: \(C=\dfrac{\left[244\left(395-195\right)\right]}{122\cdot200}=\dfrac{244}{122}=2\)
d: \(D=\dfrac{355\cdot45\cdot55}{11\cdot9\cdot7}=\dfrac{355\cdot5\cdot5}{7}=\dfrac{8875}{7}\)
1)x - 26 =95 2)79 - x = 65 3)119 - x =97 4) (x - 67) -23 =10
x =95+26 x=79-65 x =119-97 x - 67 =10+23 =33
x =121 x=14 x =22 x =33+67
x =100
231 - ( x - 6 ) = 133 : 33
231 - ( x - 6 ) = \(\frac{133}{33}\)
x - 6 = 231 - \(\frac{133}{33}\)
x - 6 = \(\frac{7490}{33}\)
x = \(\frac{7688}{33}\)
Vậy x .......
Ai thấy mình đúng tk nha !
231-( x-6)= 133:13
231- (x-6)=10,...
x-6 = 231-10, ....
x-6 = 221, ......
x = 221,...+6
x= 227, .....
133.(x+9)=1197
x+9=1197:133
x+9=9
x=9-9
x=0
133.(x+9)=1197
(x+9)=1197:133
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