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0,(57)=0,(01).57 = \(\frac{1}{99}.57=\frac{19}{33}\)
0,(42)=0,(01).42=\(\frac{1}{99}.42=\frac{14}{33}\)
=>0,(57)+0,(42)=\(\frac{19}{33}+\frac{14}{33}=\frac{33}{33}=1\)
Vậy 0,(57)+0,)42)=1
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a) 0,(5)+0,(3)+0,(1)
=\(\frac{5}{9}\)+\(\frac{1}{3}\)+\(\frac{1}{9}\)
=\(\frac{5}{9}\)+\(\frac{3}{9}\)+\(\frac{1}{9}\)
=\(\frac{5+3+1}{9}\)
=\(\frac{9}{9}\)
=1
b) 0,(57)+0,(42)
=\(\frac{19}{33}\)+\(\frac{14}{33}\)
=\(\frac{19+14}{33}\)
=\(\frac{33}{33}\)
=1
Chúc cậu học tốt •ω•
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a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
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x-y-z=0
=> x=y+z
y=x-z
-z=y-x
B=(1-z/x)(1-x/y)(1+y/z)
B=((x-z)/x)((y-x)/y)((z+y)/z)
B=(y/x)(-z/y)(x/z)
B=(-zyx)/(xyz)
B=-1
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( -3/4 )² ; ( -0,5 )² ; (9,7 )^0
= 9/16 ; = 1/4 ; = 1
( -2/5 )³ ; ( -0,5 )³
= -8/125 ; = -1/8
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=0.9![ok ok](/media/cke24/plugins/smiley/images/ok.png)
0(57) + 0.(42) = \(\frac{57}{99}+\frac{42}{99}=\frac{99}{99}=1\)