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\(Để\frac{a}{b}< \frac{a+c}{b+d}\) thì a(b+d) < b(a+c) ↔ ab + ad , ab + bc ↔ ab < bc ↔ \(\frac{a}{b}< \frac{c}{d}\)
\(Để\frac{a+c}{b+d}< \frac{c}{d}\) thì (a+c).d < (b+d).c ↔ ad + cd < bc + cd ↔ ab < bc ↔ \(\frac{a}{b}< \frac{c}{d}\)
Bài 1:
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(3-2x\right)^2=\left(x-2\right)^2\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3-x+2\right)\left(2x-3+x-2\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(3x-5\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow x=1\)
b: \(\left|x\right|< 3\)
nên -3<x<3
c: \(\left|x\right|\ge5\)
nên \(\left[{}\begin{matrix}x\ge5\\x\le-5\end{matrix}\right.\)
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=7\end{matrix}\right.\)
Bài 1:
\(a)\left(\dfrac{-28}{29}\right).\left(\dfrac{-38}{16}\right)=\dfrac{\left(-28\right).\left(-38\right)}{29.16}=\dfrac{1064}{464}=\dfrac{133}{58}\)
\(b)\left(\dfrac{-21}{16}\right).\left(\dfrac{-24}{7}\right)=\dfrac{\left(-21\right).\left(-24\right)}{16.7}=\dfrac{504}{112}=\dfrac{9}{2}\)
\(c)\left|\dfrac{-12}{17}\right|.\left(\dfrac{-34}{9}\right)=\dfrac{12}{17}.\left(\dfrac{-34}{9}\right)=\dfrac{12.\left(-34\right)}{17.9}=\dfrac{-408}{153}=\dfrac{-8}{3}\)
Bài 3:
\(a)\left|x\right|=21\)
\(\Rightarrow\left[{}\begin{matrix}x=-21\\x=21\end{matrix}\right.\)
\(b)\left|x\right|=\dfrac{17}{9};x< 0\)
\(\Rightarrow x=\dfrac{-17}{9}\)
\(c)\left|x\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\left|x\right|=\dfrac{2}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)
\(d)\left|x\right|=0,35;x>0\)
\(\Rightarrow x=0,35\)
Bài 4:
\(a)\left|x\right|-1,7=2,3\)
\(\Rightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-3}{5}\end{matrix}\right.\)
\(b)\left|x\right|+\dfrac{3}{4}-\dfrac{1}{3}=0\)
\(\Rightarrow\left|x\right|+\dfrac{3}{4}=0+\dfrac{1}{3}\)
\(\Rightarrow\left|x\right|+\dfrac{3}{4}=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=\dfrac{-1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)
Chúc bạn học tốt!