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a) \(\frac{22}{7}\div\left(11-\chi\right)=\frac{7}{5}-\frac{2}{3}\)
\(\frac{22}{7}\div\left(11-\chi\right)=\frac{11}{15}\)
\(\left(11-\chi\right)=\frac{22}{7}\div\frac{11}{15}\)
\(\left(11-\chi\right)=\frac{30}{7}\)
\(\chi=11-\frac{30}{7}\)
\(\chi=\frac{47}{7}\)
b) (x+1)+(x+2)+(x+3)+...+(x+100)=5550
Từ 1 đến 100 có 100 số hạng => Có 100 x
(x + x + x + .... + x) + (1 + 2 + 3 + .. + 100) = 5550
Áp dụng tính chất cộng dãy số cách đều, ta có
(100.x) + 5050 = 5550
100.x = 5550 - 5050
100.x = 500
x = 500 : 100
x = 5
\(71+65\cdot4=\frac{x+140}{x}+260\)
\(71+260=\frac{x+140}{x}+260\)
\(71=\frac{x}{x}+\frac{140}{x}\)
\(71=1+\frac{140}{x}\)
\(71-1=\frac{140}{x}\)
\(70=\frac{140}{x};\Rightarrow x=2\)
\(71+65\times4=\frac{x+140}{x}+260\)
\(71+260=\frac{x+140}{x}+260\)
\(71=\frac{x+140}{x}\)
\(\Rightarrow71x=x+140\)
\(\Rightarrow71x-x=140\)
\(\Rightarrow70x=140\)
\(\Rightarrow x=2\)
a) 3/x-7 = 27/135
3/x-7 = 3/15
x - 7 = 15
x = 15 + 7
x = 22
a) \(\frac{3}{x-7}=\frac{27}{135}\)
\(\Rightarrow\)\(\left(x-7\right).27=3.135\)
\(\Rightarrow\)\(\left(x-7\right).27=405\)
\(\Rightarrow\)\(x-7=15\)
\(\Rightarrow\)\(x=22\)
Vậy \(x=22\)
b ) \(71+65.4=\frac{x+140}{x}+260\)
\(71+260=\frac{x+140}{x}+260\)
\(331=\frac{x+140}{x}+260\)
\(331-260=\frac{x+140}{x}\)
\(71=\frac{x+140}{x}\)
\(71=\frac{x}{x}+\frac{140}{x}\)
\(71=1+\frac{140}{x}\)
\(70=\frac{140}{x}\)
\(x=140\div70\)
\(x=20\)
Vậy \(x=20\)
#TQY
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
a) \(0,18=0\Rightarrow x=-1\)
b)\(-\frac{14}{5}=-2,5\Rightarrow x=-3\)
a)(x.0,25+1999)x2000=(53+1999)x2000
(x.0,25+1999)=53+1999
x.0,25=53
x=53x4
x=210
a)(x.0,25+1999)x2000=(53+1999)x2000
(x.0,25+1999)=53+1999
x.0,25=53
x=53x4
x=210
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