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a) \(\frac{1,11+0,19-12,2}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
Đề đúng:
\(\frac{1,11+0,19-1,3.2^6}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
\(\frac{1,3-1,3.64}{0,39}-x=\left(\frac{3}{6}+\frac{2}{6}\right).\frac{1}{2}\)
\(\frac{1,3.\left(-63\right)}{0,39}-x=\frac{5}{6}.\frac{1}{2}\)
\(\frac{-81,9}{0,39}-x=\frac{5}{12}\)
\(\left(-210\right)-x=\frac{5}{12}\)
\(x=\left(-210\right)-\frac{5}{12}\)
\(x=\frac{-2520}{12}-\frac{5}{12}\)
\(x=\frac{-2525}{12}\)
b) \(x:\left(3\frac{1}{2}-5\frac{1}{6}\right)=4\frac{1}{5}-6\frac{2}{3}\)
\(x:\left(\frac{7}{2}-\frac{31}{6}\right)=\frac{21}{5}-\frac{20}{3}\)
\(x:\left(\frac{21}{6}-\frac{31}{6}\right)=\frac{63}{15}-\frac{100}{15}\)
\(x:\frac{-10}{6}=\frac{-37}{15}\)
\(x.\frac{6}{-10}=\frac{-37}{5}\)
\(x.\frac{-6}{10}=\frac{-37}{5}\)
\(x.\frac{-3}{5}=\frac{-37}{5}\)
\(x=\frac{-37}{5}:\frac{-3}{5}\)
\(x=\frac{-37}{5}.\frac{5}{-3}\)
\(x=\frac{-37}{-3}\)
\(x=\frac{37}{3}\)
Sai đề bài
Lẽ ra câu a chỉ chia 2 thì chia 22
Sai đề bài mina-san
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}=1\frac{9}{11}\)
=>\(\left\{1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}\right\}\times\frac{1}{2}=1\frac{9}{11}\times\frac{1}{2}\)
=>\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)
=>\(1-\frac{1}{x+1}=\frac{10}{11}\)
=> \(\frac{1}{x+1}=1-\frac{10}{11}\)
=> \(\frac{1}{x+1}=\frac{1}{11}\)
=> x + 1 = 11
=> x = 10
Nhấn đúng cho mk nha^^
a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
=> (2,85 - x - 0,5) : 0,25 = 60
=> (2,85 - 0,5) - x = 60 . 0,25
=> 2,35 - x = 15
=> x = 2,35 - 15
=> x = -12,65
Vậy x = -12,65
b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)
\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)
\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)
\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)
\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)
\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)
\(\Rightarrow x=\frac{26}{6}\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}\)
c) \(35\left(2\frac{1}{5}-x\right)=32\)
\(\Rightarrow2\frac{1}{5}-x=32\div35\)
\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)
\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)
\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)
\(\Rightarrow x=\frac{56}{405}\)
Vậy \(x=\frac{56}{405}\)
e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)
\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow4.\frac{12}{11}=11-5\div x\)
\(\Rightarrow11-5\div x=\frac{48}{11}\)
\(\Rightarrow5\div x=11-\frac{48}{11}\)
\(\Rightarrow5\div x=\frac{73}{11}\)
\(\Rightarrow x=5\div\frac{73}{11}\)
\(\Rightarrow x=\frac{55}{73}\)
Vậy \(x=\frac{55}{73}\)
a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
(2,85 - x - 0,5) : 0,25 = 60
(2,85 - x - 0,5) = 60 x 0,25
(2,85 - x - 0,5) = 15
2,35 - x = 15
x = 2,35 - 15
x = -12,65
1/2 . 1/3 . 1/4 . 1/5 . 1/6 . ( x - 1,010 ) = 1/360 - 1/720
1/2 . 1/3 . 1/4 . 1/5 . 1/6 . ( x - 1,010) = 1/720
( x - 1,010 ) . 1/2 . 1/3 . 1/4 . 1/5 . 1/6 = 1/720
( x - 1,010 ) . 1/720 = 1/720
x - 1,010 = 1/720 : 1/720
x - 1,010 = 1
x = 1 + 1,010
x = 2,01
Câu b
Ta có :x + 3 /1.3 +3/3.5 + 3/5.7+...+3/13.15=2 1/5
X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1=11/5
X+2/3.(1-1/15)=11/5
X+ 2/3.14/15=11/5
X + 28/45=11/5
X = 11/5 -28/45
X=71/45
Câu a gợi ý
1/2-1/3/1/6=0
1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) =0
3/4:x=9/10
X = 3/4:9/10
X = 5/6
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\)\(\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Rightarrow x+1=2001\)
\(\Rightarrow x=2001-1=2000\)
Vậy \(x=2000.\)
Chỗ \(x\) phải là \(\frac{2}{x\left(x+1\right)}\) chứ bạn :)
Ta có :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\) ( nhân hai vế cho \(\frac{1}{2}\) )
\(\Leftrightarrow\)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2001}\)
\(\Leftrightarrow\)\(x-1=2001\)
\(\Leftrightarrow\)\(x=2001+1\)
\(\Leftrightarrow\)\(x=2002\)
Vậy \(x=2002\)
Chúc bạn học tốt ~