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a) X : 5 + X x\(\frac{1}{10}=0,24\)
X x \(\frac{1}{5}\)+ X x \(\frac{1}{10}=0,24\)
X x \(\left(\frac{1}{5}+\frac{1}{10}\right)\)=0,24
X x \(\frac{3}{10}\)= 0 , 2 4
X =0,8
b) X : 4 - X x \(\frac{1}{100}\)= 0,12
X x \(\frac{1}{4}\)- X x \(\frac{1}{100}\)=0,12
X x \(\left(\frac{1}{4}-\frac{1}{100}\right)\)=0,12
X x \(\frac{6}{25}\)=0,12
X =0,5
a ) x : 5 + x . \(\frac{1}{10}\)= 0,24
x . \(\frac{1}{5}\)+ x . \(\frac{1}{10}\) = 0,24
x . \(\left(\frac{1}{5}+\frac{1}{10}\right)\)= 0,24
x . \(\frac{3}{10}\)= 0,24
x = 0,24 : \(\frac{3}{10}\)
x = 0,8
b ) x : 4 - x . \(\frac{1}{100}\) = 0,12
x . \(\frac{1}{4}\)- x . \(\frac{1}{100}\)= 0,12
x . \(\left(\frac{1}{4}-\frac{1}{100}\right)\)= 0,12
x . \(\frac{6}{25}\)= 0,12
x = 0,12 : \(\frac{6}{25}\)
x = 0,5
Bài 1:
Câu D
Bài 2
0,2999<0,29999;0,299999;0,2999999<3/10
=>3 giá trị của x=(0,29999;0,299999;0,2999999)
a,D
b,\(\frac{3}{10}=0,3\)
suy ra 0,2999<x<0,3 suy ra x=0,29991,x=0,29992,x=0,29993
\(\Leftrightarrow10\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}\right)=9\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=9\div10\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)
\(\Rightarrow x+1=10\)
\(\Leftrightarrow x=9\)
Vậy x = 9
Áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
VT=\(x-\left(\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-...\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{99}-\frac{1}{100}\right)\right)\)
=\(x-\frac{1}{100}\)
Dễ dàng tìm được
\(x-\frac{1}{100}=\frac{1}{100}\)
\(x=\frac{1}{50}\)
1.
( 10 - 9,34 ) + ( 10 - 9,66 )
= 0,66 + 0,34
= 1
12 - ( 12 - 9,36 )
= 12 - 2,64
= 9,36
38,52 + 0,302 x 100 - 0,72
= 38,52 + 30,2 - 0,72
= 68,72 - 0,72
= 68
2.
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0+1=1\)
\(\left(x-\frac{151}{12}\right):\frac{50}{3}=1\)
\(x-\frac{151}{12}=1\times\frac{50}{3}\)
\(x-\frac{151}{12}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{151}{12}\)
\(x=\frac{117}{4}\)
Bài 1
\(\left(10-9,34\right)+\left(10-9,66\right)\)
\(=0.66+0,34=1\)
\(12-\left(12-9,36\right)\)
\(=12-2.64=9,36\)
\(38,52+0,302.100-0,72\)
\(=38,52+30,2-0,72\)
\(=7.6\)
Bài 2
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=1\)
\(\frac{43}{8}+x-\frac{173}{24}=\frac{3}{50}\)
\(\frac{43}{8}+x=\frac{3}{50}+\frac{173}{24}\)
\(\frac{43}{8}+x=\frac{4361}{600}\)
\(x=\frac{142}{75}\)( gửi lại bài 2 đã )
\(\frac{168}{10}-x=18.65-\frac{1488}{100}\)
\(\frac{84}{5}-x=\frac{1865}{100}-\frac{1488}{100}\)
\(\frac{84}{5}-x=\frac{377}{100}\)
\(x=\frac{84}{5}-\frac{377}{100}\)
\(x=\frac{1680}{100}-\frac{377}{100}\)
\(x=\frac{1303}{100}\)
\(x=13.03\)