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Bài 4:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{c}{a}=\dfrac{d}{b}\)
hay \(\dfrac{a+c}{a}=\dfrac{b+d}{b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x-\frac{1}{3}\right)< 0\)
Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau
Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)
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\(\text{f(1)=}2.1^2+1=3\)
\(\text{f(-1)=}2.\left(-1\right)^2+1=3\)
\(\text{f(2)=}2.2^2+1=9\)
\(\text{f(0)=}2.0^2+1=1\)
\(\text{f(-3)=}=2.\left(-3\right)^2+1=19\)
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Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
\(e,\Rightarrow\dfrac{9x+7}{9}\cdot\dfrac{3}{8}-\dfrac{x}{3}=\dfrac{11}{15}-\dfrac{61}{90}=\dfrac{1}{18}\\ \Rightarrow\dfrac{9x+7}{24}-\dfrac{x}{3}=\dfrac{1}{18}\\ \Rightarrow27x+21-24x=4\\ \Rightarrow3x=-17\Rightarrow x=-\dfrac{17}{3}\)
\(f,\Rightarrow\left[{}\begin{matrix}\dfrac{1}{7}x=\dfrac{2}{7}\\\dfrac{3}{5}x=\dfrac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,\Rightarrow\left|2x+3\right|< \dfrac{6}{21}:\dfrac{3}{7}=\dfrac{2}{3}\\ \Rightarrow-\dfrac{2}{3}< 2x+3< \dfrac{2}{3}\\ \Rightarrow-\dfrac{11}{3}< 2x< -\dfrac{7}{3}\\ \Rightarrow-\dfrac{11}{6}< x< -\dfrac{7}{6}\)
\(h,\Rightarrow\left|x-\dfrac{1}{5}\right|=-\dfrac{1}{2}:2=-\dfrac{1}{4}\\ \Rightarrow x\in\varnothing\left(\left|x-\dfrac{1}{5}\right|\ge0\right)\\ i,\Rightarrow\left|x+\dfrac{1}{2}\right|=\left|\dfrac{3}{4}x+1\right|\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{3}{4}x+1\\x+\dfrac{1}{2}=-\dfrac{3}{4}x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{6}{7}\end{matrix}\right.\)
\(k,\Rightarrow\left|3x+\dfrac{2}{5}\right|=x-\dfrac{3}{5}\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{2}{5}=x-\dfrac{3}{5}\left(x\ge-\dfrac{2}{15}\right)\\3x+\dfrac{2}{5}=\dfrac{3}{5}-x\left(x< -\dfrac{2}{15}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktm\right)\\x=\dfrac{1}{20}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x\in\varnothing\)