10 x 5 + 10 x 95 = \?
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\(...\Rightarrow x+x+\dfrac{x}{43}+\dfrac{x}{8}=14+148+\dfrac{10}{30}+\dfrac{5}{95}\)
\(\Rightarrow\left(1+1+\dfrac{1}{43}+\dfrac{1}{8}\right)x=162+\dfrac{1}{3}+\dfrac{1}{19}\)
\(\Rightarrow\left(\dfrac{2.43.8}{43.8}+\dfrac{1.8}{43.8}+\dfrac{1.43}{43.8}\right)x=\dfrac{162.3.19}{3.19}+\dfrac{1.19}{3.19}+\dfrac{1.3}{19.3}\)
\(\Rightarrow\left(\dfrac{688}{344}+\dfrac{8}{344}+\dfrac{43}{344}\right)x=\dfrac{9234}{57}+\dfrac{19}{57}+\dfrac{3}{57}\)
\(\Rightarrow\dfrac{739}{344}x=\dfrac{9256}{57}\)
\(\Rightarrow x=\dfrac{9256}{57}:\dfrac{739}{344}=\dfrac{9256}{57}.\dfrac{344}{739}=\dfrac{\text{3184064}}{\text{42123}}\)
=>4x+(10/30+14/43+5/95+130/5)=0
=>4x+(85/129+495/19)=0
=>4x+65470/2451=0
=>4x=-65470/2451
=>x=-32735/4902
a. 12x5+12x95=12x(5+95)=12x100=1200
b. 10x20x5x10=(10x10)x(20x5)=100x100=10000
tk nha
a, 12.5+12.95=12.(5+95)
=12.100
=1200
b,10.20.5.10=(10.10).(20.5)
=100.100
=10000
Answer:
Có:
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
\(\Rightarrow\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x+100}{8}-6=0\)
\(\Rightarrow\left(\frac{x-10}{30}-3\right)+\left(\frac{x-14}{43}-2\right)+\left(\frac{x-5}{95}-1\right)+\frac{x-100}{8}=0\)
\(\Rightarrow\frac{x-100}{30}+\frac{x-100}{43}+\frac{x-100}{95}+\frac{x-100}{8}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\ne0\Leftrightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
a, 200 - 3( x - 16 ) = 20
3( x - 16 ) = 200 - 20 = 180
x - 16 = 180 : 3 = 60
x = 60 + 16 = 76
b, 5 + 10 + 15 + .............. + 95 + 100 + 105 = 1200
c, x + ( 99 - 97 + 95 - 93 + ............ + 7 - 5 + 3 - 1 ) = 100
x + ( 2 . 25 ) = 100
x + 50 = 100
x = 100 - 50 = 50
****, thks
\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)
=>x-100=0
hay x=100
10 x 5 + 10 x 95 = 10 x ( 5 + 95 )
= 10 x 100
= 1000
1000
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