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\(2,\frac{9}{x}=\frac{2}{5}-\frac{7}{20}\)
\(\Rightarrow\frac{9}{x}=\frac{1}{20}\)
\(\Rightarrow x=9.20\)
\(\Rightarrow x=180\)
\(\frac{x}{5}=\frac{5}{6}+\left(-\frac{19}{30}\right)\)
\(\frac{\Rightarrow x}{5}=\frac{1}{5}\)
\(\Rightarrow x=1\)
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Mình ko ghi lai de
C=3.(1-1/20+1/13-1/2013)/7.(1-1/20+1/13-1/2013)
C=3.1/7.1
C=3/7
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a ) 5x=2y
b) 7x = 3y
c ) 3+x /7+y= 3/7
=> (3+x).7= (7+y).3
21 + 7x = 21 + 3y
=> 7x = 3y
=> x/y = 3/7 => x= 20:(7+3).3=6
=> y = 20 - 6 = 14
Vậy (x;y) thuộc {(6;14)}
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\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)
\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)
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l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
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a) \(\frac{-2}{x}=\frac{y}{5}\Leftrightarrow xy=-10\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-10\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy : ...
b) \(\frac{3+x}{7+y}=\frac{3}{7}\)
\(\Leftrightarrow21+7x=21+3y\)
\(\Leftrightarrow7x=3y\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\) ( Tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=14\end{matrix}\right.\)
Vậy : ....
c) \(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{x+y}{5+3}=\frac{16}{8}=2\) ( Tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\left\{{}\begin{matrix}x=10\\y=6\end{matrix}\right.\)
Vậy : ...
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a) Ta có: x chẵn
⇔x=2k(k∈N)
⇔\(x^2=4k^2\)(1)
Ta có: y chẵn
⇔y=2b(b∈N)
⇔\(y^2=4b^2\)(2)
Cộng (1) và (2) ta được: \(x^2+y^2=4k^2+4b^2=4\left(k^2+b^2\right)⋮4\)(đpcm)
b) Ta có: x⋮7
⇔x=7a(a∈N)
⇔\(x^2=49a^2\)(3)
Ta có: y⋮7
⇔y=7c
⇔\(y^2=49c^2\)(4)
Cộng (3) và (4) ta được: \(x^2+y^2=49a^2+49c^2=49\left(a^2+c^2\right)⋮7\)(đpcm)
5y2 +7y + 20 ⋮ y-2
⇔ 5y2 - 10y + 17y - 34 + 54 ⋮ y-2
⇔ 5y(y-2) + 17 (y - 2) +54 ⋮ y-2
⇔ 54 ⋮ y-2
⇔ y-2 ϵ {1 ; 2;3;6; 9; 18;27;54}
y ϵ {3;4;5;8;11;20;29;56}