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a, \(B=\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\dfrac{2^{10}.\left(13+65\right)}{2^8.2^3.13}\)
\(=\dfrac{2^{10}.78}{2^{11}.13}\)\(=\dfrac{1.6}{2.1}=\dfrac{1.3}{1.1}=3\)
b: \(=\dfrac{2^{20}\cdot3^2+2^{54}}{2^{18}\cdot5^2}=\dfrac{2^{20}\left(3^2+2^{32}\right)}{2^{18}\cdot5^2}=\dfrac{2^2\left(3^2+2^{32}\right)}{25}\)
c: \(=\dfrac{2^9\cdot3^6\cdot3^6\cdot2^2}{2^8\cdot3^{12}}=\dfrac{2^{11}}{2^8}=8\)
d: \(=\dfrac{2^{12}\cdot3^4\cdot3^{10}}{2^{12}\cdot3^{12}}=9\)
\(\dfrac{72^3.54^2}{108^4}\)
\(=\dfrac{18^3.4^3.18^2.3^2}{18^4.6^4}\)
\(=\dfrac{18^5.2^6.3^2}{18^4.2^4.3^4}\)
\(=\dfrac{18.2^2}{3^2}\)
\(=\dfrac{9.2.2^2}{9}\)
\(=2^3=8\)
a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
Theo đề : \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2+y^2+2z^2=108\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\left(\dfrac{z}{4}\right)^2\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=2.\left(\dfrac{z}{4}\right)^2=>\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2+y^2+2z^2}{4+9+32}=\dfrac{108}{45}=\dfrac{12}{5}\)
Với \(\dfrac{x^2}{2}=\dfrac{12}{5}\Rightarrow x^2=\dfrac{12}{5}.2=\dfrac{24}{5}\Rightarrow x=\dfrac{2\sqrt{30}}{5}\)
\(\dfrac{y^2}{3}=\dfrac{12}{5}\Rightarrow y^2=\dfrac{12}{5}.3=\dfrac{36}{5}\Rightarrow y=\dfrac{6\sqrt{5}}{5}\)
\(\dfrac{2z^2}{4}=\dfrac{12}{5}\Rightarrow2z^2=\dfrac{12}{5}.4=\dfrac{48}{5}\Rightarrow z^2=\dfrac{24}{5}=>\dfrac{2\sqrt{30}}{5}\)
theo bài ra ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
dựa vào tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\ =\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\\ =\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)=>x=4.2=8
=>y=4.3=12 =>z=4.4=16 vậy x,y,z lần lượt là 8;12;16
Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\Rightarrow x=2k;y=\dfrac{3k}{2};z=\dfrac{4k}{3}\)
Khi đó \(xyz=-108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=-108\)
\(\dfrac{24k^3}{6}=-108\Leftrightarrow4k^3=-108\)
\(\Leftrightarrow k^3=-27\Rightarrow k=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=\dfrac{3k}{2}=\dfrac{3\cdot\left(-3\right)}{2}=-\dfrac{9}{2}\\z=\dfrac{4k}{3}=\dfrac{4\cdot\left(-3\right)}{3}=-4\end{matrix}\right.\)
\(\dfrac{2}{3}+\dfrac{2}{54}+\dfrac{2}{108}+....+\dfrac{2}{648}\\ =\dfrac{2}{3}+\dfrac{2}{6.9}+\dfrac{2}{9.12}+...+\dfrac{2}{24.27}\\ =\dfrac{2}{3}+\dfrac{2}{3}\left(\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{24.27}\right)\\ =\dfrac{2}{3}+\dfrac{2}{3}.\left(\dfrac{1}{6}-\dfrac{1}{27}\right)\\ =\dfrac{2}{3}+\dfrac{2}{3}.\dfrac{7}{54}\\ =\dfrac{2}{3}+\dfrac{7}{81}\\ =\dfrac{61}{81}\)
Ta có: x/2=y/3=z/4→x2/4=y2/9=2x2/32=x2-y2+2z2/4-9+32=108/27=81
Với x/2=81→x=81.2=162
Với y/3=81→y=81.3=243
Với z/4=81→81.4=324
Ta có:\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{2z^2}{32}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\dfrac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=4\)
\(\dfrac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow y=6\)
\(\dfrac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow z=8\)
=8
mk hỏi cách làm chứ ko nhờ bn bấm máy tính