Tính :
\(\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
^Mk cần câu TL rõ ràng nhé
Những câu hỏi liên quan
Tính giá trị của các biểu thức sau:
a) \(A=\frac{5x2^{13}x4^{11}-16^9}{\left(3x2^{17}\right)^2}\)
b) \(B=\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
a)\(A=\frac{5.2^{13}.2^{22}-2^{36}}{\left(3.2^{17}\right)^2}\)
\(A=\frac{5.2^{35}-2^{36}}{3^2.2^{34}}\)
\(A=\frac{2^{35}\left(5-2\right)}{3^2.2^{34}}\)
\(A=\frac{2.3}{3^2}=\frac{2}{3}\)
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b) \(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(2+3\right)}\)
\(B=\frac{7}{2.5}=\frac{7}{10}\)
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a) (\(\frac{-1}{2}\))3- (\(\frac{3}{4}\))3x(-2)2
b) 2x(-1)5+(\(\frac{3}{4}\))2-\(\frac{3}{8}\)
c) \(\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
ính
(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
Help ikkk, tui like cho
(69.210.210) : (219.273+15.49.94)
=((2.3)9.(2.2)10): (219.(33)3+15.(22)9.(32)4)
= (29.39.(22)10) : (219.39+15.218.38)
= (29.39.220) : (218.38.(2.3+15))
= (29.39.22.218) :(218.38.21)
=(211.39.218) : (218.39.7)
=211:7
=\(\frac{2048}{7}\)
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Bài 1:
a) A=2^19x27^3+15x4^9x9/6^9x2^10+1^210
Các bạn giải giúp mik nhé, nếu sai đề thì bảo mik, mik cảm ơn^^
(69x210+1210):(219x273+15x49x94) = ?
4 tính gia tri bieu thuc
M=219x273+15x49x94
69x210+1210
Tính x+y+z nếu :
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{7x}{y+z}+\frac{7y}{x+y}+\frac{7z}{x+y}=\frac{133}{10}\)\(\frac{133}{10}\)
Cần Câu TL rõ Ràng
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)
\(\Rightarrow19\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{133}{10}\)
\(\Rightarrow\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{7}{10}\)
\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)
\(\Rightarrow7\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{133}{10}\)
\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{19}{10}\)
\(\Rightarrow\left(\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\right)=\frac{19}{10}+3\)
\(\Rightarrow\left(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\right)=\frac{49}{10}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}\)
\(\Rightarrow\left(x+y+z\right).\frac{7}{10}=\frac{49}{10}\)
\(\Rightarrow x+y+z=7\)
Vậy x + y + z = 7