Gọi độ dài tấm vải thứ nhất, hai, ba lần lượt là a,b,c(m)(145>a,b,c>0)
Ta có: \(\left(1-\dfrac{1}{2}\right)a=\left(1-\dfrac{1}{3}\right)b=\left(1-\dfrac{1}{4}\right)c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{3}=\dfrac{3c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a+b+c}{2+\dfrac{3}{2}+\dfrac{4}{3}}=\dfrac{145}{\dfrac{29}{6}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}a=30.2=60\\b=\dfrac{3}{2}.30=45\\c=\dfrac{4}{3}.30=40\end{matrix}\right.\)(nhận)
Vậy...
Bài 97:
\(\Leftrightarrow x-68=\left(\dfrac{43}{11}-4\right):\left(\dfrac{21}{22}-1\right):\left(\dfrac{33}{34}-1\right)=-68\Leftrightarrow x=0\)
Bài 98:
a) \(A=\dfrac{2002}{\left|x\right|+2003}\le\dfrac{2002}{0+2003}=\dfrac{2002}{2003}\)
\(maxA=2003\Leftrightarrow x=0\)
b) \(B=\dfrac{\left|x\right|+2002}{-2003}\le\dfrac{0+2002}{-2003}=-\dfrac{2002}{2003}\)
\(maxB=-\dfrac{2002}{2003}\Leftrightarrow x=0\)
