87- (73-x) =20
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\(87-\left(73-x\right)=20\)
\(73-x=87-20\)
\(73-x=67\)
\(x=73-67\)
\(x=6\)
\(87-\left(73-x\right)=20\\ \Rightarrow73-x=67\\ \Rightarrow x=6.\)
\(\dfrac{x-13}{87}+\dfrac{x-27}{73}+\dfrac{x-67}{83}+\dfrac{x-73}{27}=4\)
<=>\(\dfrac{x-13}{87}-1+\dfrac{x-27}{73}-1+\dfrac{x-67}{83}-1+\dfrac{x-73}{27}-1=0\)
<=>\(\dfrac{x-100}{87}+\dfrac{x-100}{73}+\dfrac{x-100}{83}+\dfrac{x-100}{27}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}\right)=0\)
Do \(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}>0\)
=>x-100=0
<=>x=100
25 x 33 + 25 x 87 + 64 x 73 + 64 x 47
= 25 x ( 33 + 87 ) + 64 x ( 73 + 47 )
= 25 x 120 + 64 x 120
= 120 x ( 15 + 64 )
= 120 x 49
= 5580
25 x 33 + 25 x 87 + 64 x 73 + 64 x 47
= 25 x ( 33 + 87 ) + 64 x ( 73 +47 )
= 25 x 120 + 64 x 120
= 120 x ( 25 + 64 )
= 120 x 89
= 10680
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
(x-13)/87+(x-27)/73+(x-67)/33+(x-73)/27=4
=>(x-13-87)/87+(x-27-73)/73+(x-67-33)/33+(x-73-27)/27=4-1-1-1-1
=>(x-100)/87+(x-100)/73+(x-100)/33+(x-100)/27=0
=>(x-100)*(1/87+1/73+1/33+1/27)=0
=>x-100=0
=>x=100
Ta có :
\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)
\(\Leftrightarrow x=100\)
Vậy \(x=100\)
\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)
\(\Rightarrow x-100=0\Leftrightarrow x=100\)
Lời giải:
$87-(73-x)=20$
$73-x=87-20=67$
$x=73-67=6$
87 - (73 - \(x\)) = 20
87 - 73+ \(x\) = 20
14+ \(x\) = 20
\(x\) = 20 - 14
\(x\) = 6