y x 5 x2 +4 +6 +...+18=270
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y.5+(2+4+6+...+18) = 270
y.5+(18+2).9:2=270
y.5+90=270
y.5=180
y=36
Trước tiên ta tính tổng 2 + 4 + 6 + ...... + 18
Số các số hạng là (18 - 2) : 2 + 1 = 9 (số)
Tổng là: 9 x (2 + 18) = 180
y x 5 + 180 = 270
y x 5 = 270 - 180 = 90
y = 90 : 5 = 18
Ta có: x:y:z =4:5:6
⇒\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}\)
⇒\(\dfrac{x^2}{16}=\dfrac{2y^2}{50}=\dfrac{z^2}{36}\)
⇒\(\dfrac{x^2-2y^2+z^2}{16-50+36}=\dfrac{18}{2}=9\)
\(\dfrac{x}{4}=9\Rightarrow x=36\)
\(\dfrac{y}{5}=9\Rightarrow y=45\)
\(\dfrac{z}{6}=9\Rightarrow z=54\)
Y x 5 + 2 + 4 + 6 + ..... + 18 = 270
Y x 5 + ( 2 + 4 + 6 + ..... + 18 ) = 270
Y x 5 + ( ( 2 + 18 ) + ( 4 + 16 ) + ( 6 + 14 ) + ( 8 + 12 ) + 10 ) = 270
Y x 5 + ( 20 + 20 + 20 + 20 + 10 ) = 270
Y x 5 + ( 20 x 4 + 10 ) = 270
Y x 5 + 80 = 270
Y x 5 = 270 - 80
Y x 5 = 190
Y = 190 : 5
Y = 39
y x 5 + (2 + 4 + 6 +...+ 18) = 270
=> y x 5 + (18 + 2) x 9 : 2 = 270 (Công thức tính tổng)
=> y x 5 + 20 x 9 : 2 = 270
=> y x 5 + 180 : 2 = 270
=> y x 5 + 90 = 270
=> y x 5 = 270 - 90
=> y x 5 = 180
=> y = 180 : 5
=> y = 36
Vậy...
a: \(\frac{x+1}{2x-6}-\frac{4}{2x-6}\)
\(=\frac{x+1-4}{2\left(x-3\right)}\)
\(=\frac{x-3}{2\left(x-3\right)}=\frac12\)
b: \(\frac{3x-4}{6x+3}-\frac{x-5}{6x+3}\)
\(=\frac{3x-4-x+5}{6x+3}\)
\(=\frac{2x+1}{3\left(2x+1\right)}=\frac13\)
c: \(\frac{x-1}{x-3}-\frac{3x-8}{3-x}+\frac{3-2x}{x-3}\)
\(=\frac{x-1}{x-3}+\frac{3x-8}{x-3}+\frac{3-2x}{x-3}\)
\(=\frac{x-1+3x-8+3-2x}{x-3}=\frac{2x-6}{x-3}=\frac{2\left(x-3\right)}{x-3}\)
=2
d: \(\frac{3}{x+5}-\frac{5}{x-7}\)
\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)
\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)
e: \(\frac{3}{x+5}-\frac{5}{x-7}\)
\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)
\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)
f: \(\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{x^2-4}\)
\(=\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2\left(x+2\right)+3\left(x-2\right)+5x-18}{\left(x-2\right)\left(x+2\right)}=\frac{2x+4+3x-6+5x-18}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{10x-20}{\left(x-2\right)\left(x+2\right)}=\frac{10}{x+2}\)
18.2
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18,2
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