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29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29 = -5
1 + 29-x/21 + 1 + 27-x/23 + 1 + 25-x/25 + 1 + 23-x/27 + 1 + 21-x/29 = 0
50-x/21 + 50-x/23 + 50-x/25 + 50-x/27 + 50-x/29 = 0
(50-x) (1/21 + 1/23 + 1/25 + 1/27 + 1/29) = 0
Vì: 1/21 + 1/23 + 1/25 + 1/27 + 1/2 > 0
=> 50 - x = 0
x = 50
Vậy x = 50
\(\frac{-1}{3}+\frac{0,2-0,3+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{10}+\frac{5}{11}}{\frac{-3}{10}+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{39}{110}}{\frac{-79}{80}}\)
\(=\frac{-1}{3}-\frac{312}{869}\)
\(=\frac{-1805}{2607}\)
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
1a, 34.26.15−34.44.15=34.15.(26−44)=320.(−18)=−271034.26.15−34.44.15=34.15.(26−44)=320.(−18)=−2710
b,−√25+2√16=−√27.4=−4√27−25+216=−27.4=−427
2a, 2+x=-5
=>x=-7
b, x−1x+5=67=>7(x−1)=6(x+5)=>7x−7=6x+30=>x=37
1a, \(\frac{3}{4}.26.\frac{1}{5}-\frac{3}{4}.44.\frac{1}{5}=\frac{3}{4}.\frac{1}{5}.\left(26-44\right)=\frac{3}{20}.\left(-18\right)=\frac{-27}{10}\)
b,\(-\sqrt{25+2}\sqrt{16}=-\sqrt{27}.4=-4\sqrt{27}\)
2a, 2+x=-5
=>x=-7
b, \(\frac{x-1}{x+5}=\frac{6}{7}\\ =>7\left(x-1\right)=6\left(x+5\right)\\ =>7x-7=6x+30\\ =>x=37\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)
b)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Vậy \(x=-329\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
a
\(6,5-\frac{9}{4}:\left|x+\frac{1}{3}\right|=2\)
\(\Leftrightarrow\frac{9}{4}:\left|x+\frac{1}{3}\right|=4,5\)
\(\Leftrightarrow\frac{9}{4}:\frac{9}{2}=\left|x+\frac{1}{3}\right|\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=\frac{1}{2}\)
\(TH1:x+\frac{1}{3}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
\(TH1:x+\frac{1}{3}=-\frac{1}{2}\)
\(\Leftrightarrow x=-\frac{5}{6}\)
b
\(x^{2020}-x^{2018}=0\)
\(\Leftrightarrow x^{2018}\left(x^2-1\right)=0\)
\(\Leftrightarrow x^{2018}\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow x=1;x=-1;x=0\)
1. Chú ý này: 1+27=3+25=5+23=...=25+3=27+1=28
Giải:
Đặt:
\(C=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{25}+\frac{1}{27}\)
\(A=\frac{1}{1.27}+\frac{1}{3.25}+\frac{1}{5.23}+...+\frac{1}{13.15}+\frac{1}{15.13}+...+\frac{1}{25.3}+\frac{1}{27.1}\)
=> \(A=2\left(\frac{1}{1.27}+\frac{1}{3.25}+\frac{1}{5.23}+...+\frac{1}{13.15}\right)\)
=> \(14.A=28\left(\frac{1}{1.27}+\frac{1}{3.25}+\frac{1}{5.23}+...+\frac{1}{13.15}\right)\)
\(=\frac{28}{1.27}+\frac{28}{3.25}+\frac{28}{5.23}+...+\frac{28}{13.15}\)
\(=\frac{1+27}{1.27}+\frac{3+25}{3.25}+\frac{5+23}{5.23}+...+\frac{13+15}{13.15}\)
\(=1+\frac{1}{27}+\frac{1}{3}+\frac{1}{25}+\frac{1}{5}+\frac{1}{23}+...+\frac{1}{13}+\frac{1}{15}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}\)
=> \(A=\frac{1}{14}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{27}\right)=\frac{1}{14}C\)=> C=14A
=> \(B=C:A\)=14A:A=14
Ta có: x−227+x−326+x−425+x−445=0x−227+x−326+x−425+x−445=0
⇔x−2927+x−2926+x−2925+x−295=0⇔x−2927+x−2926+x−2925+x−295=0
⇔x−29=0⇔x−29=0
hay x=29
đúng hong = )
\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-44}{5}=0\)
<=> \(\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\frac{x-29}{5}=0\)
<=> \(\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{5}=0\)
<=> \(\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\right)=0\)
<=> x - 29 = 0 (vì \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\ne0\))
<=> x = 29
Vậy x = 29