-(\(\dfrac{-1}{6}\))\(^{100}\) và -(\(\dfrac{-1}{8}\))\(^{150}\)
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\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)
Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)
Lấy (4) trừ (3) ta có:
\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)
=>\(B=\dfrac{32+16+6+2+1}{64}\)
=>\(B=\dfrac{63}{64}\)
\(\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(0,5-1\dfrac{3}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(\dfrac{1}{2}-\dfrac{8}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\dfrac{11}{10}\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{33}{80}\)
\(\Rightarrow x:2,2=\dfrac{33}{80}:\dfrac{1}{6}\)
\(\Rightarrow x:2,2=\dfrac{99}{40}\)
\(\Rightarrow x=\dfrac{99}{40}\times2,2\)
\(\Rightarrow x=\dfrac{1089}{200}\)
=>(x:2,2)*1/6=-3/8(1/2-8/5)=33/80
=>x:2,2=99/40
=>x=1089/200
Ta có:
\(-\left(-\dfrac{1}{6}\right)^{100}=-\left(\dfrac{1}{6}\right)^{100}=-\left[\left(\dfrac{1}{6}\right)^2\right]^{50}=-\left(\dfrac{1}{36}\right)^{50}\)
\(-\left(-\dfrac{1}{8}\right)^{150}=-\left(\dfrac{1}{8}\right)^{150}=-\left[\left(\dfrac{1}{8}\right)^3\right]^{50}=-\left(\dfrac{1}{512}\right)^{50}\)
Mà: \(\dfrac{1}{36}>\dfrac{1}{512}\)
\(\Rightarrow\left(\dfrac{1}{36}\right)^{50}>\left(\dfrac{1}{512}\right)^{50}\)
\(\Rightarrow-\left(\dfrac{1}{36}\right)^{50}< -\left(\dfrac{1}{512}\right)^{50}\)
\(\Rightarrow-\left(\dfrac{-1}{6}\right)^{100}< -\left(\dfrac{-1}{8}\right)^{150}\)