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a)\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)
\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)
\(\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)
\(\frac{1}{3}.x=\frac{35}{12}\)
\(x=\frac{35}{12}:\frac{1}{3}\)
\(x=\frac{35}{12}.\frac{3}{1}\)
\(x=\frac{35}{4}\)
b)\(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)
\(4,5.\left(0,1.x\right)=2,25.0,3\)
\(4,5.\left(0,1.x\right)=0,675\)
\(0,1.x=0,675:4,5\)
\(0,1.x=0,15\)
\(x=0,15:0,1\)
\(x=1.5\)
c)\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)
\(\left(\frac{1}{4}.x\right).2=8.0,02\)
\(\left(\frac{1}{4}.x\right).2=0,16\)
\(\frac{1}{4}.x=0,16:2\)
\(\frac{1}{4}.x=0,08\)
\(x=0,08:\frac{1}{4}\)
\(x=0,32\)
d)\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)
\(3.\left(6.x\right)=\frac{3}{4}.\frac{9}{4}\)
\(3.\left(6.x\right)=\frac{27}{16}\)
\(6.x=\frac{27}{16}:3\)
\(6.x=\frac{9}{16}\)
\(x=\frac{9}{16}:6\)
\(x=\frac{3}{32}\)
HỌC TỐT ^^
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)
<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
<=> x = 2010