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31−43−(−53)+721−92−361+151
=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=31−43+53+721−92−361+151
=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(-\frac{3}{4}-\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{72}=(31−92)+(−43−361)+(53+151)+721
=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(-\frac{27}{36}-\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\frac{1}{72}=(93−92)+(−3627−361)+(159+151)+721
=\frac{1}{9}+\frac{-7}{9}+\frac{2}{3}+\frac{1}{72}=91+9−7+32+721
=-\frac{2}{3}+\frac{2}{3}+\frac{1}{72}=−32+32+721
=0+\frac{1}{72}=\frac{1}{72}=0+721=721
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\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
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Có: \(\frac{y-2}{3}=\frac{2y-4}{6}\)
\(\frac{z-3}{4}=\frac{3z-9}{12}\)
Suy ra\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)
\(=\frac{\left(x-2y+3z\right)-6}{8}=\frac{14-6}{8}=1\)
Vậy có \(\frac{x-1}{2};\frac{y-2}{3};\frac{z-3}{4}=1\)Thay vào có x=3; y=5; z=7
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Ta co x+2/2=y+3/3=z+4/4 hay x+1=y+1=z+1 => x=y=z
Suy ra: 2x+y+z=11 hay 2x+x+x=11 => 4x=11 => x=11/4
Vay: x^2+y^2+z^2 = (11/4)^2+(11/4)^2+(11/4)^2 =121/16 . 3 = 363/16
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Vì (x-2)^2=9
(x-2)^2 = 3^2
Nên x-2=3
hoặc x-2=-3
=> x=5 hoặc x=-1
\(\left(x-2\right)^2=9\)
\(\Leftrightarrow\left(x-2\right)^2=3^2\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5.\)
Vậy \(x=5.\)
#Riin
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\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)
\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)
\(\Rightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)
\(\Rightarrow3x=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{7}{4}:3\)
\(\Rightarrow x=\dfrac{7}{12}\)
\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
\(d,4^{x-3}+1=17\)
\(\Rightarrow4^{x-3}=17-1\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
#Toru
`3/2 x -1 =1/2x -3/5`
`=> 3/2x -1/2x = -3/5 +1`
`=> 2/2x= -3/5 + 5/5`
`=> x= 2/5`
__
`1/2x +1/2(x-2) = 3/4 -2x`
`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`
`=> 1/2x +1/2x +2x = 3/4 + 1`
`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`
`=> 6/2x = 7/4`
`=> x= 7/4 : 3`
`=>x=7/12`
__
`(x-1/2) -1/4=0`
`=> x-1/2=1/4`
`=> x=1/4 +1/2`
`=> x= 1/4 +2/4`
`=>x=3/4`
__
`4^(x-3) +1=17`
`=> 4^(x-3) =17-1`
`=> 4^(x-3)=16`
`=> 4^(x-3)=4^2`
`=> x-3=2`
`=>x=2+3`
`=>x=5`
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a) \(\left|x-1,7\right|+\dfrac{1}{2}=23\)
\(\left|x-1,7\right|=22,5\)
\(\Rightarrow x-1,7=\pm22,5\)
\(\Rightarrow x=\left\{24,2;-20,8\right\}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75\)
\(\left|x+\dfrac{4}{15}\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=\pm1,6\)
\(\Rightarrow x=\left\{\dfrac{4}{3};-\dfrac{28}{15}\right\}\)
x3:x=4/9
=> x2=4/9
=> \(x=\frac{-2}{3}\) hoặc \(x=\frac{2}{3}\)
x^3:x = 4/9
=> x^2 = 4/9
=> x^2 = {(+-2)/(+-3)}^2
=> x = 2/3 hoặc -2/-3