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\(\dfrac{1}{x\times5}=\dfrac{121}{27830}\)
\(\dfrac{1}{x\times5}=\dfrac{1}{230}\)
\(x\times5=230\)
\(x=230:5\)
\(x=46\)
\(\dfrac{1}{x\cdot5}=\dfrac{121}{27830}\)
\(x\cdot5=27830:121\)
\(x\cdot5=230\)
\(x=230:5\)
\(x=46\)
Mình không hiểu yêu cầu đề bài lắm, bạn có thể viết rõ ra được không ạ
2 x X - ( 67+ 34) :2 = 121
2 x X - 101 : 2 = 121
2 x X = 121 + 101 : 2
2 x X = 121 + 50,5
2 x X = 171,5
X = 171,5 : 2
X = 85,75
\(598+\left(123-121\right)x2\)
\(=598+2x2\)
\(=598+4\)
\(=602\)
`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{14}{9}\div\dfrac{2}{121}\right)\times\dfrac{3}{77}\)
`=`\(\left(\dfrac{14}{9}\times\dfrac{121}{2}\right)\times\dfrac{3}{77}\)
`=`\(\left(\dfrac{7}{9}\times121\right)\times\dfrac{3}{77}\)
`=`\(\dfrac{847}{9}\times\dfrac{3}{77}=\dfrac{11}{3}\times1=\dfrac{11}{3}\)
(14/9:2/121) x 3/77
= ( 7/9 x 121/) x 3/77
= 847/9 x 3/77
=11/3 x 1=11/3
`#3107.101107`
a)
\(x+x+\dfrac{1}{2}\times\dfrac{2}{5}+x+\dfrac{8}{10}=121\\3x+\dfrac{1}{5}+\dfrac{4}{5}=121\\ 3x+1=121\\ 3x=121-1\\ 3x=120\\ x=40 \)
Vậy, `x = 40`
b)
\(\dfrac{12+x}{42}=\dfrac{5}{6}\\ \dfrac{12+x}{42}=\dfrac{35}{42}\\ \dfrac{12+x}{42}-\dfrac{35}{42}=0\\ \dfrac{12+x-35}{42}=0\\ \dfrac{x-\left(35-12\right)}{42}=0\\ \dfrac{x-23}{42}=0\\ x-23=0\\ x=23\)
Vậy,` x = 23.`
a: \(x+x+\dfrac{1}{2}\cdot\dfrac{2}{5}+x+\dfrac{8}{10}=121\)
=>\(3x+\dfrac{1}{5}+\dfrac{4}{5}=121\)
=>3x+1=121
=>3x=120
=>x=40
b: \(\dfrac{x+12}{42}=\dfrac{5}{6}\)
=>\(x+12=42\cdot\dfrac{5}{6}=35\)
=>x=35-12=23
(1-1/4).(1-1/9).(1-1/16)....(1-1/100).(1-1/121)
=3/4.8/9.15/16...99/100.120/121
=(1.3/2.2).(2.4/3.3).(3.5/4.4)....(9.11/10.10).(10.12/11.11)
=1.3.2.4.3.5...9.11.10.12/2.2.3.3.4.4...10.10.11.11
=(2.3.4...9.10.11).(3.4.5...10.12)/(2.3.4...9.10.11).(2.3.4....10.11)
=12/2.11
=6/11
nha
\(121:\left(2\times x-1\right)=121\\ 2\times x-1=121:121\\ 2\times x-1=1\\ 2\times x=1+1\\ 2\times x=2\\ x=2:2\\ x=1\)