(0,175)^0-9^1009*1/3^2018+(1/4)^3:1/2^8
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2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
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\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)
\(=\frac{11}{16}:\frac{11}{32}\)
\(=\frac{11}{16}.\frac{32}{11}\)
\(=2\)
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Ta có : \(\overline{abcabc}:\overline{abc}=1001\)
\(\Rightarrow\) \(\overline{abcabc}=\overline{abc}\times1001=\overline{abc}\times7\times11\times13\)
Vậy \(\overline{abcabc}\)là bội của 7; 11; 13
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Tìm x :
a/ \(x+17=-33\)
\(x=-33-17\)
\(x=-50\)
b/ \(2-\left(x-5\right)=5\times2^3\)
\(2-\left(x-5\right)=40\)
\(x-5=2-40\)
\(x-5=-38\)
\(x=-38+5\)
c/ \(1009.x=\left(-1\right)+2+\left(-3\right)+4+\left(-5\right)+6+....+\left(-2017\right)+2018\)
\(1009.x=\left(-1\right)+\left(-1\right)+\left(-1\right)+....+\left(-1\right)\)
\(1009.x=\left(-1\right).1010\)
\(1009.x=-1009\)
\(\Rightarrow x=-1\)
\(x=-33\)
1, ta có : abcabc=1000abc +abc=1001abc chia hết cho 7,11,13
2,a,x+17=-33 b,2-(x-5)=5.2^3 c,1009.x=-1+2+(-3)+4+...+(-2017)+2018
x=-33-17 2-(x-5)=5.8 1009.x=(-1+2)+(-3+4)+...+(-2017+2018)
x=-50 2-(x-5)=40 1009x=1+1+1+...+1+1
x-5=-38 1009.x=1009.1
x=-33 1009.x=1009
x=1
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Ta có: \(x^2+y^2=1\Leftrightarrow\left(x^2+y^2\right)^2=1\) (1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right)ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1009}=\left(\frac{y^2}{b}\right)^{1009}=\left(\frac{1}{a+b}\right)^{1009}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}=\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}+\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}+\frac{1}{\left(a+b\right)^{1009}}=\frac{2}{\left(a+b\right)^{1009}}\left(đpcm\right)\)
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A = 0-1 + 2-3 + 4-5 +...+ 2017-2018
=> A = (-1) + (-1) + (-1) +...+ (-1) (Có 1009 số hạng)
=> A = 1009.(-1)
=> A = -1009
B = 1-3+5-7+ 9-11+....+2005-2007
=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)
=> B = 502.(-2)
=> B = -1004
C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102
=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)
=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)
=> C = (-9).17
=> C = -153