(2/3)^3 .3 (-3/4)^2 . (-1)2003 // (2/5)^2. (-5/12)^3
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LT
2
KB
15 tháng 11 2016
\(\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\frac{\frac{8}{27}.\frac{9}{16}.\left(-1\right)}{\frac{4}{25}.\frac{-125}{1728}}=\frac{\frac{-72}{432}}{\frac{-500}{43200}}=\frac{\frac{-1}{6}}{\frac{-5}{432}}=\frac{-1}{6}.\frac{432}{-5}=14,4\)
27 tháng 7 2017
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)
Bài này dễ mà bn
3 tháng 5 2021
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
3 tháng 5 2021
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\left(\frac{2}{3}\right)^3.\left(\frac{3}{4}\right)^2.\left(-1\right)}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\left(\frac{2}{3}\right)^3.\left(\frac{3}{4}\right)^2}{\left(\frac{2}{5}\right)^2.\left(\frac{5}{12}\right)^3}\)
\(=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}}{\frac{2^2}{5^2}.\frac{5^3}{12^3}}\)
\(=\frac{\frac{2^3.3^2}{3^3.4^2}}{\frac{2^2.5^3}{5^2.12^3}}\)
\(=\frac{2^3.3^2.5^2.12^3}{3^3.4^2.2^2.5^3}\)
\(=\frac{2^3.3^2.5^2.\left(4.3\right)^3}{3^3.\left(2^2\right)^2.2^2.5^3}\)
\(=\frac{2^3.3^2.5^2.4^3.3^3}{3^3.2^4.2^2.5^3}\)
\(=\frac{2^3.3^2.5^2.\left(2^2\right)^3.3^3}{3^3.2^4.2^2.5^3}\)
\(=\frac{2^3.3^2.5^2.2^6.3^3}{3^3.2^4.2^2.5^3}\)
\(=\frac{2^9.3^5.5^2}{3^3.2^6.5^3}\)
\(=\frac{2^3.3^2}{5}\)
\(=\frac{8.9}{5}\)
\(=\frac{72}{5}.\)
Chúc bạn học tốt!