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1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)
Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)
\(\left(y+\frac{3}{7}\right)^{468}=0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)
\(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)
=> \(x-\frac{2}{5}=0\)
\(y-\frac{3}{7}=0\)
=> \(x=\frac{2}{5}\)
\(y=\frac{3}{7}\)
Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
\(\left(x+1\right)+\left(x+4\right)+...+\left(x+28\right)=155155\)
\(=>\left(10x\right)+\frac{\left(28+1\right).10}{2}=155155\)
\(=>10x+\frac{290}{2}=155155\)
\(=>10x=155155-\frac{290}{2}=155155-145\)
\(=>10x=155010\)
\(=>x=\frac{155010}{10}=15510\)
( 7.3 + 8.13) : ( 9/2/3 - y) = 39
( 21 + 104) : ( 29/3 - y) = 39
125 : ( 29/3-y) = 39
29/3-y = 125 - 39
29/3-y = 86
y = 29/3 -86
y = -229/3
b, (y+1) + (y+2) + (y+3) + .... + (y+99) = 6138
(y+y+y+y+....+y) + (1+2+3+...+99) = 6138
99.y + 4950 = 6138
99.y = 6138 - 4950
99.y = 1188
y = 1188 : 99
x = 12
a) \(\left(y+1\right)+\left(y+4\right)+\left(y+7\right)+....+\left(y+28\right)=155155\)
\(\Rightarrow\left(y+y+...+y\right)+\left(1+4+7+...+28\right)=155155\)
\(\Rightarrow10x+145=155155\)
\(\Rightarrow10x=155010\)
\(\Rightarrow x=15501\)
Vậy x = 15501
b) \(\left(y+1\right)+\left(y+2\right)+..+\left(y+99\right)=6138\)
\(\Rightarrow\left(y+y+...+y\right)+\left(1+2+3+...+99\right)=6138\)
\(\Rightarrow99x+4950=6138\)
\(\Rightarrow99x=1188\)
\(\Rightarrow x=12\)
Vậy x = 12