120 x 4 + 120 x 5 + 120
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\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne30\\x\ne24\end{cases}}\)
Ta có \(\frac{60}{\frac{120}{x}-4}+\frac{60}{\frac{120}{x}-5}=x\)
\(\Leftrightarrow\frac{60}{\frac{120-4x}{x}}+\frac{60}{\frac{120-5x}{x}}=x\)
\(\Leftrightarrow\frac{60x}{120-4x}+\frac{60x}{120-5x}=x\)
\(\Leftrightarrow\frac{60}{120-4x}+\frac{60}{120-5x}=1\left(Do\text{ }x\ne0\right)\)
\(\Leftrightarrow\frac{15}{30-x}=1-\frac{12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{24-x-12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{12-x}{24-x}\)
\(\Leftrightarrow360-15x=\left(12-x\right)\left(30-x\right)\)
\(\Leftrightarrow360-15x=360-42x+x^2\)
\(\Leftrightarrow x^2-27x=0\)
\(\Leftrightarrow x\left(x-27\right)=0\)
\(\Leftrightarrow x=27\left(Tm\text{ }ĐKXĐ\right)\)
\(\frac{120\times999+120}{2\times4\times0,8\times5\times2,5\times125}\)
\(=\frac{120\times1000}{\left(125\times0,8\right)\left(4\times2,5\right)\left(2\times5\right)}\)
\(=\frac{120000}{100\times10\times10}\)
\(=\frac{120000}{10000}\)
\(=\frac{12}{1}=12\)
TL
a) 5x + 20 = 110
<=> x = 90 : 5 = 18
b) x + 18 = - 13
<=> x = - 31
c) 120 - x = 50
<=> x = 70
d) 10 - x = -29
<=> x = 39
e) - x + 31 = 61
<=> x = -30
f) -85 - x = --70
<=> x = 15
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ĐK: ` x \ne 10; x \ne 0`
`120/(x-10)-3/5=120/x`
`<=>120/(x-10)-120/x=3/5`
`<=>1/(x-10) - 1/x= 1/200`
`<=> (x-x+10)/(x(x-10)) = 1/200`
`<=> 10/(x(x-10))= 1/200`
`<=> x^2-10=2000`
`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy `S={50;-40}`.
`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`
`<=>40/(x-10)-1/5=40/x`
`<=>200x-x(x-10)=200(x-10)`
`<=>200x-200x+2000-x^2+10x=0`
`<=>x^2-10x-2000=0`
`Delta'=25+2000=2025`
`<=>x_1=50,x_2=-40`
Vậy `S={50,-40}`
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)
\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)
\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)
\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)
\(\Leftrightarrow-3x^2+1230x-6000=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)
Vậy ...
\(5x-40:4=120\)
\(\Leftrightarrow5x-40=120\).4
\(\Leftrightarrow5x=480+40\)
\(\Leftrightarrow x=520:5\)
\(\Leftrightarrow x=104\)
A, 120 x 2 = 120 x 4
120 x 2 = 240
B, số đó là :
(2270 - 1034) : 4 = 309
= 120 x 4 + 120 x 5 + 120 x 1
=120 x (4+5+1)
=120 x 10
=1200
120 x 4 + 120 x 5+ 120
=120 x 4 + 120 x 5+ 120 + 1
=120 x(4+5+1)
=120 x 10
=1200