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x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ... + ( x + 100 ) = 11 000
<=> ( x + x + x + x + x + ... + x ) + ( 1 + 2 + 3 + 4 + ... + 100 ) = 11 000
<=> 101x + \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 11 000 ( vì sao em để 101x thì idol biết mà :33 )
<=> 101x + 5050 = 11 000
<=> 101x = 5950
<=> x = 5950/101
Mình ghi nhầm đó, lẽ ra phải thế này cơ:\(\left(30:7\frac{1}{2}+0,5x3-1,5\right)x\left(4\frac{1}{2}-\frac{9}{2}\right):\left(14,5x100\right)\)
\((30:7\frac{1}{2}\)\(+0,5x3\)\(-1,5)x(4\frac{1}{2}\)\(-\frac{9}{2}\)\():(14,5x100)\)
\(=\)\((30:7\frac{1}{2}\)\(+0,5x3\) \(-1,5)x(\frac{9}{2}\)\(-\frac{9}{2}\)\()\)\(:(14,5x100)\)
\(=30:7\frac{1}{2}\)\(+0,5x3-1,5)x0x(14,5x100)\)
\(=0\)
Chúc bạn học tốt!
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
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\)\(\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Rightarrow x+1=2001\)
\(\Rightarrow x=2001-1=2000\)
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 ~
Bài làm:
Xét: \(101\times102-101\times101-50-51=101\times\left(102-101\right)-101=101\times1-101=0\)
\(\Rightarrow\left(1+2+4+8+...+512\right)\times\left(101\times102-101\times101-50-51\right)=0\)
\(\Rightarrow\frac{\left(1+2+4+8+...+512\right)\times\left(101\times102-101\times101-50-51\right)}{2+4+8+16+...+1024+2048}=0\)
Học tốt!!!!
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}=1\frac{9}{11}\)
=>\(\left\{1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\times\left(x+1\right)}\right\}\times\frac{1}{2}=1\frac{9}{11}\times\frac{1}{2}\)
=>\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{10}{11}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{x}-\frac{1}{x+1}=\frac{10}{11}\)
=>\(1-\frac{1}{x+1}=\frac{10}{11}\)
=> \(\frac{1}{x+1}=1-\frac{10}{11}\)
=> \(\frac{1}{x+1}=\frac{1}{11}\)
=> x + 1 = 11
=> x = 10
Nhấn đúng cho mk nha^^
Ta có:(x + 1)+(x+3)+...+(x+9)=75
=>5x + (1+3+...+9)=75
=>5x + 25=75
=>5x =50
=>x = 10
a. (x+1)+(x+3)+(x+5)+(x+7)+(x+9)=75
=>(x+x+x+x+x)+(1+3+5+7+9)=75
=>Xx5+25=75
=>Xx5=75-25
=>Xx5=50
=>x=50:5
=>x=10
b. sai đề
36x + (1+2+3+...+36) =810
36x +(1+36)x 36:2 =810
36x+ 666= 810
36x =144
x=4
Vậy x=4
Từ 1 đến 36 có 36 số tự nhiên liên tiếp . do đó tổng đã cho có 36 chữ X
Tổng 36 số tự nhiên liên tiếp từ 1 đến 36 là:
(1 + 36) x 36 : 2 = 666
Tổng các nhóm đã cho là:
X x 36 + 666 = 810
X x 36 =810 - 666
X x 36 = 144
X =144 : 36
X =4