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Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}<\frac{1}{11^{21}}\) Vì 2733 > 1133 > 1121
a: \(A=\left|x-4,1\right|+\left|x-\frac23\right|-9\)
\(=\left|x-\frac23\right|+\left|x-4,1\right|-9\)
TH1: x<2/3
=>\(x-\frac23<0;x-4,1<0\)
=>\(A=\frac23-x+4,1-x-9=-4,9+\frac23-2x=\frac{-49}{10}+\frac23-2x=\frac{-147+20}{30}-2x=\frac{-127}{30}-2x\)
TH2: \(\frac23\le x<4,1\)
=>\(x-\frac23\ge0;x-4,1<0\)
=>\(A=x-\frac23+4,1-x-9=-\frac23+4,1-9=-\frac23-4.9=-\frac23-\frac{49}{10}=\frac{-20-147}{30}=-\frac{167}{30}\)
TH3: x>=4,1
=>\(x-\frac23>0;x-4,1\ge0\)
=>\(A=x-\frac23+x-4,1-9=2x-\frac23-13,1=2x-\frac23-\frac{131}{10}=2x-\frac{20+393}{30}=2x-\frac{413}{30}\)
b: \(B=\left|2\frac15-x\right|+\left|x-\frac15\right|+8\frac15\)
\(=\left|x-0,2\right|+\left|x-2,2\right|+8,2\)
TH1: x<0,2
=>x-0,2<0; x-2,2<0
=>B=0,2-x+2,2-x+8,2=10,6-2x
TH2: 0,2<=x<2,2
=>x-0,2>=0; x-2,2<0
=>B=x-0,2+2,2-x+8,2=2+8,2=10,2
TH3: x>=2,2
=>x-0,2>0; x-2,2>=0
=>B=x-0,2+x-2,2+8,2=2x+5,8
c: \(C=\left|x+3\frac12\right|+\left|x\right|-3\frac12\)
\(=\left|x+3,5\right|+\left|x\right|-3,5\)
TH1: x<-3,5
=>x+3,5<0; x<0
=>C=-x-3,5-x-3,5=-2x-7
TH2: -3,5<=x<0
=>x+3,5>=0; x<0
=>C=x+3,5-x-3,5=0
TH3: x>=0
=>x+3,5>0; x>=0
=>C=x+3,5+x-3,5=2x